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Latihan Soal Perhitungan Kimia Sederhana (Mudah)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Aluminium berekasi dengan asam klorida menghasilkan larutan aluminium klorida dan gas hidrogen.
$\mbox{Al}(s)\mbox{+4HNO}_{3}(aq)$$\rightarrow\mbox{Al}\left(\mbox{NO}_{3}\right)_{3}(aq)$$+\mbox{NO}(g)\mbox{+2H}_{2}\mbox{O}(l)$
Jika 2 mol $\mbox{HNO}_{3}$ bereaksi dengan aluminium, berapa mol gas $\mbox{NO}$ yang dihasilkan?
Betul$\begin{array}{ccc}
\mbox{mol NO} & : & \mbox{mol HNO}_{3}\\
1 & : & 4\\
\mbox{mol NO} & = & \frac{1}{4}\times2\mbox{ mol}\\
& = & 0,5\mbox{ mol}
\end{array}$Salah$\begin{array}{ccc}
\mbox{mol NO} & : & \mbox{mol HNO}_{3}\\
1 & : & 4\\
\mbox{mol NO} & = & \frac{1}{4}\times2\mbox{ mol}\\
& = & 0,5\mbox{ mol}
\end{array}$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
2 mol besi bereaksi dengan gas klorin $\left(\mbox{Cl}_{2}\right)$ berdasarkan reaksi berikut:
$3\mbox{Fe}(s)\mbox{+Cl}_{2}(g)$$\rightarrow\mbox{3FeCl}_{2}(s)$
Volume gas klorin pada kondisi RTP yang dibutuhkan agar besi habis beraksi adalah … .
Betul$\begin{array}{ccc}
\mbox{mol Fe} & : & \mbox{mol Cl}_{2}\\
3 & : & 1
\end{array}$$\begin{alignedat}{1}\mbox{mol Cl}_{2} & =\frac{1}{3}\times\mbox{mol Fe}\\
& =\frac{1}{3}\times2\mbox{ mol}\\
& =\frac{2}{3}\mbox{ mol}\\
\mbox{Volume Cl}_{2} & =\mbox{mol Cl}_{2}\times V_{M}\\
& =\frac{2}{3}\times24\mbox{ dm}^{3}\\
& =16\mbox{ dm}^{3}.
\end{alignedat}
$Salah$\begin{array}{ccc}
\mbox{mol Fe} & : & \mbox{mol Cl}_{2}\\
3 & : & 1
\end{array}$$\begin{alignedat}{1}\mbox{mol Cl}_{2} & =\frac{1}{3}\times\mbox{mol Fe}\\
& =\frac{1}{3}\times2\mbox{ mol}\\
& =\frac{2}{3}\mbox{ mol}\\
\mbox{Volume Cl}_{2} & =\mbox{mol Cl}_{2}\times V_{M}\\
& =\frac{2}{3}\times24\mbox{ dm}^{3}\\
& =16\mbox{ dm}^{3}.
\end{alignedat}
$ -
Pertanyaan ke 3 dari 5
3. Pertanyaan
10 gram batu kapur $\left(\mbox{CaCO}_{3}\right)$ direaksikan dengan asam klorida menghasilkan gas karbon dioksida. Berapakah massa maksimum karbon dioksida yang dihasilkan?
($A_{r}$ C = 12; O = 16; Ca = 40; H = 1; Cl = 35,5)
$\mbox{CaCO}_{3}+2\mbox{HCl}$$\rightarrow\mbox{CaCl}_{2}+\mbox{H}_{2}\mbox{O}+\mbox{CO}_{2}.$
Betul$\mbox{CaCO}_{3}+2\mbox{HCl}$$\rightarrow\mbox{CaCl}_{2}+\mbox{H}_{2}\mbox{O}+\mbox{CO}_{2}$
massa $\mbox{CaCO}_{3}=10$ gram
$\begin{aligned}\mbox{mol CaCO}_{3} & =\frac{\mbox{massa CaCO}_{3}}{M_{r}\mbox{ CaCO}_{3}}\\
& =\frac{10}{100}\\
& =0,1\mbox{ mol}
\end{aligned}
$$\mbox{molCaCO}_{3}\mbox{: mol CO}_{2}$$=1:\,1$
$\begin{aligned}\mbox{mol CO}_{2} & =\frac{1}{1}\times\mbox{mol CaCO}_{3}\\
& =\frac{1}{1}\times0,1\mbox{ mol}\\
& =0,1\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{massa CO}_{2} & =\mbox{mol CO}_{2}\times M_{r}\mbox{ CO}_{2}\\
& =0,1\times44\\
& =4,4\mbox{ gram}.
\end{aligned}
$Salah$\mbox{CaCO}_{3}+2\mbox{HCl}$$\rightarrow\mbox{CaCl}_{2}+\mbox{H}_{2}\mbox{O}+\mbox{CO}_{2}$
massa $\mbox{CaCO}_{3}=10$ gram
$\begin{aligned}\mbox{mol CaCO}_{3} & =\frac{\mbox{massa CaCO}_{3}}{M_{r}\mbox{ CaCO}_{3}}\\
& =\frac{10}{100}\\
& =0,1\mbox{ mol}
\end{aligned}
$$\mbox{molCaCO}_{3}\mbox{: mol CO}_{2}$$=1:\,1$
$\begin{aligned}\mbox{mol CO}_{2} & =\frac{1}{1}\times\mbox{mol CaCO}_{3}\\
& =\frac{1}{1}\times0,1\mbox{ mol}\\
& =0,1\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{massa CO}_{2} & =\mbox{mol CO}_{2}\times M_{r}\mbox{ CO}_{2}\\
& =0,1\times44\\
& =4,4\mbox{ gram}.
\end{aligned}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
Pada RTP, 6 $\mbox{dm}^{3}$gas hidrogen dihasilkan dari reaksi seng dengan 200 mL asam klorida, sebagai berikut:
$\mbox{Zn+2HCl}\rightarrow\mbox{ZnCl}_{2}+\mbox{H}_{2}$
Konsentrasi asam klorida adalah … .
Betul$\mbox{Zn+2HCl}\rightarrow\mbox{ZnCl}_{2}+\mbox{H}_{2}$
Volume $\mbox{H}_{2}$= 6 $\mbox{dm}^{3}$
Volume HCl = 200 mL = 0,2 L
$\begin{alignedat}{1}\mbox{mol H}_{2} & =\frac{\mbox{volume gas}}{\mbox{volume molar}}\\
& =\frac{6}{24}\\
& =0,25\mbox{ mol}\\
\mbox{mol H}_{2} & :\mbox{ mol HCl}\\
1 & :\,2\\
\mbox{mol HCl} & =\frac{2}{1}\times\mbox{mol H}_{2}\\
& =2\times0,25\\
& =0,5\mbox{ mol}\\
M & =\frac{n}{v}\\
& =\frac{0,5}{0,2}\\
& =2,5\mbox{ M.}
\end{alignedat}
$Salah$\mbox{Zn+2HCl}\rightarrow\mbox{ZnCl}_{2}+\mbox{H}_{2}$
Volume $\mbox{H}_{2}$= 6 $\mbox{dm}^{3}$
Volume HCl = 200 mL = 0,2 L
$\begin{alignedat}{1}\mbox{mol H}_{2} & =\frac{\mbox{volume gas}}{\mbox{volume molar}}\\
& =\frac{6}{24}\\
& =0,25\mbox{ mol}\\
\mbox{mol H}_{2} & :\mbox{ mol HCl}\\
1 & :\,2\\
\mbox{mol HCl} & =\frac{2}{1}\times\mbox{mol H}_{2}\\
& =2\times0,25\\
& =0,5\mbox{ mol}\\
M & =\frac{n}{v}\\
& =\frac{0,5}{0,2}\\
& =2,5\mbox{ M.}
\end{alignedat}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Pada titrasi asam-basa, 20 mL larutan asam sulfat $\left(\mbox{H}_{2}\mbox{SO}_{4}\right)$ dinetralkan oleh 23 mL larutan natrium hidroksida (NaOH) 0,1 M. Berapakah konsentrasi asam sulfat tersebut ?
Betul$\begin{array}{c}
\mbox{2NaOH}\\
2
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}\mbox{SO}_{4}\\
: & 1
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Na}_{2}\mbox{SO}_{4}\\
: & 1
\end{array}$ $\begin{array}{cc}
+ & \mbox{2H}_{2}\mbox{O}\\
: & 2
\end{array}$V NaOH = 23 mL = 0,023 L
M NaOH = 0,1 M
V $\mbox{H}_{2}\mbox{SO}_{4}$= 20 mL = 0,02 L
$\begin{alignedat}{1}\mbox{mol NaOH} & =M\times V\\
& =0,1\times0,023\\
& =0,0023\mbox{ mol}\\
\mbox{mol H}_{2}\mbox{SO}_{4} & =\frac{1}{2}\times0,0023\mbox{ mol}\\
& =1,15\times10^{-3}\mbox{mol}\\
\mbox{M H}_{2}\mbox{SO}_{4} & =\frac{n}{V}\\
& =\frac{1,15\times10^{-3}}{0,02}\mbox{ M}\\
& =0,0575\mbox{ M}.
\end{alignedat}
$Salah$\begin{array}{c}
\mbox{2NaOH}\\
2
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}\mbox{SO}_{4}\\
: & 1
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Na}_{2}\mbox{SO}_{4}\\
: & 1
\end{array}$ $\begin{array}{cc}
+ & \mbox{2H}_{2}\mbox{O}\\
: & 2
\end{array}$V NaOH = 23 mL = 0,023 L
M NaOH = 0,1 M
V $\mbox{H}_{2}\mbox{SO}_{4}$= 20 mL = 0,02 L
$\begin{alignedat}{1}\mbox{mol NaOH} & =M\times V\\
& =0,1\times0,023\\
& =0,0023\mbox{ mol}\\
\mbox{mol H}_{2}\mbox{SO}_{4} & =\frac{1}{2}\times0,0023\mbox{ mol}\\
& =1,15\times10^{-3}\mbox{mol}\\
\mbox{M H}_{2}\mbox{SO}_{4} & =\frac{n}{V}\\
& =\frac{1,15\times10^{-3}}{0,02}\mbox{ M}\\
& =0,0575\mbox{ M}.
\end{alignedat}
$
Latihan Soal Perhitungan Kimia Sederhana (Sedang)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
3,35 gram gas etuna bereaksi di udara menghasilkan karbon dioksida dan air. Jika volume oksigen di udara adalah 20%, berapakah volume udara untuk membakar sempurna gas etuna tersebut?
BetulMassa gas $\mbox{C}_{2}\mbox{H}_{2}$ = 3,35 gram
$\begin{aligned}M_{r}\mbox{ C}_{2}\mbox{H}_{2} & =\left(2\times A_{r}\mbox{ C}\right)+\left(2\times A_{r}\mbox{ H}\right)\\
& =(2\times12)+(2\times1)\\
& =24+2\\
& =26
\end{aligned}
$$\begin{alignedat}{1}\mbox{mol C}_{2}\mbox{H}_{2} & =\frac{\mbox{massa C}_{2}\mbox{H}_{2}}{M_{r}\mbox{ C}_{2}\mbox{H}_{2}}\\
& =\frac{3,35}{26}\\
& =0,125\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
2\mbox{C}_{2}\mbox{H}_{2}\\
2
\end{array}$ $\begin{array}{cc}
+ & 5\mbox{O}_{2}\\
: & 5
\end{array}$ $\begin{array}{cc}
\rightarrow & 4\mbox{CO}_{2}\\
: & 4
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
: & 2
\end{array}$$\begin{aligned}\mbox{mol O}_{2} & =\frac{5}{2}\times\mbox{mol C}_{2}\mbox{H}_{2}\\
& =\frac{5}{2}\times0,125\mbox{ mol}\\
& =0,3125\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{volume O}_{2} & =\mbox{mol O}_{2}\times V_{M}\\
& =0,3125\times22,4\mbox{ liter}\\
& =7\mbox{ liter}
\end{aligned}
$$\begin{aligned}\%\mbox{ O}_{2} & =\frac{\mbox{volume O}_{2}}{\mbox{volume udara}}\times100\%\\
20\% & =\frac{\mbox{7 liter}}{\mbox{volume udara}}\times100\%\\
\mbox{volume udara} & =\frac{7\times100\%}{20\%}\\
& =35\mbox{ liter}.
\end{aligned}
$SalahMassa gas $\mbox{C}_{2}\mbox{H}_{2}$ = 3,35 gram
$\begin{aligned}M_{r}\mbox{ C}_{2}\mbox{H}_{2} & =\left(2\times A_{r}\mbox{ C}\right)+\left(2\times A_{r}\mbox{ H}\right)\\
& =(2\times12)+(2\times1)\\
& =24+2\\
& =26
\end{aligned}
$$\begin{alignedat}{1}\mbox{mol C}_{2}\mbox{H}_{2} & =\frac{\mbox{massa C}_{2}\mbox{H}_{2}}{M_{r}\mbox{ C}_{2}\mbox{H}_{2}}\\
& =\frac{3,35}{26}\\
& =0,125\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
2\mbox{C}_{2}\mbox{H}_{2}\\
2
\end{array}$ $\begin{array}{cc}
+ & 5\mbox{O}_{2}\\
: & 5
\end{array}$ $\begin{array}{cc}
\rightarrow & 4\mbox{CO}_{2}\\
: & 4
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
: & 2
\end{array}$$\begin{aligned}\mbox{mol O}_{2} & =\frac{5}{2}\times\mbox{mol C}_{2}\mbox{H}_{2}\\
& =\frac{5}{2}\times0,125\mbox{ mol}\\
& =0,3125\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{volume O}_{2} & =\mbox{mol O}_{2}\times V_{M}\\
& =0,3125\times22,4\mbox{ liter}\\
& =7\mbox{ liter}
\end{aligned}
$$\begin{aligned}\%\mbox{ O}_{2} & =\frac{\mbox{volume O}_{2}}{\mbox{volume udara}}\times100\%\\
20\% & =\frac{\mbox{7 liter}}{\mbox{volume udara}}\times100\%\\
\mbox{volume udara} & =\frac{7\times100\%}{20\%}\\
& =35\mbox{ liter}.
\end{aligned}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
Untuk mendapatkan logam besi murni bisa dilakukan dengan cara mereaksikan haematite $\left(\mbox{Fe}_{2}\mbox{O}_{3}\right)$ dengan karbon (C), berdasarkan reaksi berikut:
$2\mbox{Fe}_{2}\mbox{O}_{3}\mbox{+3C}$$\rightarrow\mbox{4Fe+3CO}_{2}$
Pada akhir reaksi diperoleh 720,0 $\mbox{cm}^{3}$ (RTP) dan 2,8 gram sisa haematite. Berapakah massa mula-mula $\mbox{Fe}_{2}\mbox{O}_{3}$ ?
BetulVolume $\mbox{CO}_{2}$$=720,0$ $\mbox{cm}^{3}$$=0,72$ $\mbox{dm}^{3}$
$\begin{alignedat}{1}\mbox{mol CO}_{2} & =\frac{\mbox{volume CO}_{2}}{V_{M}}\\
& =\frac{0,72\mbox{ dm}^{3}}{24\mbox{ dm}^{3}}\\
& =0,02\mbox{ mol}
\end{alignedat}
$$\begin{array}{ccccccc}
2\mbox{Fe}_{2}\mbox{O}_{3} & + & \mbox{3C} & \rightarrow & \mbox{24Fe} & + & \mbox{3CO}_{2}\\
2 & : & 3 & : & 4 & : & 3
\end{array}$$\begin{aligned}\mbox{mol Fe}_{2}\mbox{O}_{3}\mbox{reaksi} & =\frac{2}{3}\times\mbox{mol CO}_{2}\\
& =\frac{2}{3}\times0,03\mbox{ mol}\\
& =0,02\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{massa}\mbox{ Fe}_{2}\mbox{O}_{3}\mbox{reaksi} & =\mbox{mol Fe}_{2}\mbox{O}_{3}\times M_{r}\mbox{ Fe}_{2}\mbox{O}_{3}\\
& =0,02\times160\mbox{ gram}\\
& =3,2\mbox{ gram}
\end{aligned}
$$\mbox{massa Fe}_{2}\mbox{O}_{3}\mbox{ mula-mula}$ $=\mbox{massa FeCO}_{3}\mbox{ reaksi}$
$+\mbox{massa }\mbox{FeCO}_{3}\mbox{sisa}$
$\begin{aligned} & =3,2+2,8\mbox{ gram}\\
& =6\mbox{ gram}.
\end{aligned}
$SalahVolume $\mbox{CO}_{2}$$=720,0$ $\mbox{cm}^{3}$$=0,72$ $\mbox{dm}^{3}$
$\begin{alignedat}{1}\mbox{mol CO}_{2} & =\frac{\mbox{volume CO}_{2}}{V_{M}}\\
& =\frac{0,72\mbox{ dm}^{3}}{24\mbox{ dm}^{3}}\\
& =0,02\mbox{ mol}
\end{alignedat}
$$\begin{array}{ccccccc}
2\mbox{Fe}_{2}\mbox{O}_{3} & + & \mbox{3C} & \rightarrow & \mbox{24Fe} & + & \mbox{3CO}_{2}\\
2 & : & 3 & : & 4 & : & 3
\end{array}$$\begin{aligned}\mbox{mol Fe}_{2}\mbox{O}_{3}\mbox{reaksi} & =\frac{2}{3}\times\mbox{mol CO}_{2}\\
& =\frac{2}{3}\times0,03\mbox{ mol}\\
& =0,02\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{massa}\mbox{ Fe}_{2}\mbox{O}_{3}\mbox{reaksi} & =\mbox{mol Fe}_{2}\mbox{O}_{3}\times M_{r}\mbox{ Fe}_{2}\mbox{O}_{3}\\
& =0,02\times160\mbox{ gram}\\
& =3,2\mbox{ gram}
\end{aligned}
$$\mbox{massa Fe}_{2}\mbox{O}_{3}\mbox{ mula-mula}$ $=\mbox{massa FeCO}_{3}\mbox{ reaksi}$
$+\mbox{massa }\mbox{FeCO}_{3}\mbox{sisa}$
$\begin{aligned} & =3,2+2,8\mbox{ gram}\\
& =6\mbox{ gram}.
\end{aligned}
$ -
Pertanyaan ke 3 dari 5
3. Pertanyaan
4,1 gram padatan garam kalsium nitrat dipanaskan sehingga terurai menjadi kalsium oksida, gas nitrogen dioksida dan gas oksigen. Selanjutnya gas hasil reaksi di alirkan ke larutan natrium hidroksida. Berapakah volume gas yang tersisa?
Betul$\begin{array}{c}
2\mbox{Ca}\left(\mbox{NO}_{3}\right)_{2}(s)\\
2
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CaO}(s)\\
: & 2
\end{array}$$\begin{array}{cc}
+ & 4\mbox{NO}_{2}(g)\\
: & 4
\end{array}$$\begin{array}{cc}
+ & \mbox{O}_{2}(g)\\
: & 1
\end{array}$$\begin{alignedat}{1}n\mbox{ Ca}\left(\mbox{NO}_{3}\right)_{2} & =\frac{4,1}{164}\mbox{ mol}\\
& =0,025\mbox{ mol}\\
n\mbox{ NO}_{2} & =\frac{4}{2}\times0,025\mbox{ mol}\\
& =0,05\mbox{ mol}\\
n\mbox{ O}_{2} & =\frac{1}{2}\times0,025\mbox{ mol}\\
& =0,0125\mbox{ mol}
\end{alignedat}
$Gas $\mbox{NO}_{2}$ bersifat asam sehingga bisa bereaksi dengan NaOH sedangkan $\mbox{O}_{2}$ tidak bereaksi. Jadi, gas yang tersisa setelah dialirkan ke dalam larutan NaOH adalah gas oksigen sejumlah $0,0125\mbox{ mol}\times22,4\mbox{ dm}^{3}/\mbox{mol}=0,28\mbox{ dm}^{3}.$
Salah$\begin{array}{c}
2\mbox{Ca}\left(\mbox{NO}_{3}\right)_{2}(s)\\
2
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CaO}(s)\\
: & 2
\end{array}$$\begin{array}{cc}
+ & 4\mbox{NO}_{2}(g)\\
: & 4
\end{array}$$\begin{array}{cc}
+ & \mbox{O}_{2}(g)\\
: & 1
\end{array}$$\begin{alignedat}{1}n\mbox{ Ca}\left(\mbox{NO}_{3}\right)_{2} & =\frac{4,1}{164}\mbox{ mol}\\
& =0,025\mbox{ mol}\\
n\mbox{ NO}_{2} & =\frac{4}{2}\times0,025\mbox{ mol}\\
& =0,05\mbox{ mol}\\
n\mbox{ O}_{2} & =\frac{1}{2}\times0,025\mbox{ mol}\\
& =0,0125\mbox{ mol}
\end{alignedat}
$Gas $\mbox{NO}_{2}$ bersifat asam sehingga bisa bereaksi dengan NaOH sedangkan $\mbox{O}_{2}$ tidak bereaksi. Jadi, gas yang tersisa setelah dialirkan ke dalam larutan NaOH adalah gas oksigen sejumlah $0,0125\mbox{ mol}\times22,4\mbox{ dm}^{3}/\mbox{mol}=0,28\mbox{ dm}^{3}.$
-
Pertanyaan ke 4 dari 5
4. Pertanyaan
12,61 gram XS bereaksi dengan oksigen pada kondisi standar berdasarkan reaksi berikut:
$\mbox{XS+O}_{2}$$\rightarrow\mbox{ZnO+SO}_{2}$
Jika dihasilkan 2,912 $\mbox{dm}^{3}$gas $\mbox{SO}_{2}$, tentukan $A_{r}$ logam X. ($A_{r}$ S = 32; O = 16).
Betul$\begin{alignedat}{1}n_{\mbox{SO}_{2}} & =\frac{2,912}{22,4}\mbox{ mol}\\
& =0,13\mbox{ mol}
\end{alignedat}
$$\begin{array}{ccc}
\mbox{XS} & + & \mbox{O}_{2}\\
12,61\mbox{ g}\\
0,13\mbox{ mol}
\end{array}$$\begin{array}{cccc}
\rightarrow & \mbox{ZnO} & + & \mbox{SO}_{2}\\
& & & 22,4\mbox{ dm}^{3}\\
& & & 0,13\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{XS}} & =n_{\mbox{SO}_{2}}\\
& =0,13\mbox{ mol}\\
n_{\mbox{XS}} & =\frac{m_{\mbox{XS}}}{M_{r}\mbox{ XS}}\\
0,13 & =\frac{12,61}{\left(A_{r}\mbox{ X}+32\right)}\\
A_{r}\mbox{ X}+32 & =\frac{12,61}{0,13}\\
& =97\\
A_{r}\mbox{X} & =97-32\\
& =65.
\end{alignedat}
$Salah$\begin{alignedat}{1}n_{\mbox{SO}_{2}} & =\frac{2,912}{22,4}\mbox{ mol}\\
& =0,13\mbox{ mol}
\end{alignedat}
$$\begin{array}{ccc}
\mbox{XS} & + & \mbox{O}_{2}\\
12,61\mbox{ g}\\
0,13\mbox{ mol}
\end{array}$$\begin{array}{cccc}
\rightarrow & \mbox{ZnO} & + & \mbox{SO}_{2}\\
& & & 22,4\mbox{ dm}^{3}\\
& & & 0,13\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{XS}} & =n_{\mbox{SO}_{2}}\\
& =0,13\mbox{ mol}\\
n_{\mbox{XS}} & =\frac{m_{\mbox{XS}}}{M_{r}\mbox{ XS}}\\
0,13 & =\frac{12,61}{\left(A_{r}\mbox{ X}+32\right)}\\
A_{r}\mbox{ X}+32 & =\frac{12,61}{0,13}\\
& =97\\
A_{r}\mbox{X} & =97-32\\
& =65.
\end{alignedat}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Natrium nitrida biasa digunakan untuk kantung udara pada mobil. Kantung udara akan segera terisi gas seketika dengan adanya benturan keras pada bagian depan mobil. Hal ini terjadi karena natrium nitrida mengalami dekomposisi berdasarkan reaksi berikut.
$\mbox{2NaN}_{3}$$\rightarrow\mbox{2Na+3N}_{2}$
$\mbox{10Na+2KNO}_{3}$$\rightarrow\mbox{K}_{2}\mbox{O}+5\mbox{Na}_{2}\mbox{O}\mbox{+N}_{2}$
Berapakah volume gas nitrogen yang terbentuk dari 65 gram $\mbox{NaN}_{3}$?
($A_{r}$ Na = 23; N = 14; K = 39; O = 16).
Betul$\begin{alignedat}{1}m_{\mbox{NaN}_{3}} & =\frac{65}{65}\\
& =1\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
2\mbox{NaN}_{3}\\
1\mbox{ mol}\\
10\mbox{ Na}\\
1\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\rightarrow & 2\mbox{Na}\\
& 1\mbox{ mol}\\
+ & 2\mbox{KNO}_{3}\\
& 0,2\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & 3\mbox{N}_{2}\\
& 1,5\mbox{ mol}\\
\rightarrow & \mbox{K}_{2}\mbox{O}\\
& 0,1\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\\
\\
+ & 5\mbox{Na}_{2}\mbox{O}\\
& 0,5\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\\
\\
+ & \mbox{N}_{2}\\
& 0,1\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{total N}_{2}} & =1,5\mbox{ mol}+0,1\mbox{ mol}\\
& =1,6\mbox{ mol}.
\end{alignedat}
$Salah$\begin{alignedat}{1}m_{\mbox{NaN}_{3}} & =\frac{65}{65}\\
& =1\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
2\mbox{NaN}_{3}\\
1\mbox{ mol}\\
10\mbox{ Na}\\
1\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\rightarrow & 2\mbox{Na}\\
& 1\mbox{ mol}\\
+ & 2\mbox{KNO}_{3}\\
& 0,2\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & 3\mbox{N}_{2}\\
& 1,5\mbox{ mol}\\
\rightarrow & \mbox{K}_{2}\mbox{O}\\
& 0,1\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\\
\\
+ & 5\mbox{Na}_{2}\mbox{O}\\
& 0,5\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\\
\\
+ & \mbox{N}_{2}\\
& 0,1\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{total N}_{2}} & =1,5\mbox{ mol}+0,1\mbox{ mol}\\
& =1,6\mbox{ mol}.
\end{alignedat}
$
Latihan Soal Perhitungan Kimia Sederhana (Sukar)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
1,95 gram logam seng tepat bereaksi dengan larutan besi ($x$) klorida menghasilkan seng klorida dan 1,12 gram padatan besi. Berapakah nilai $x$?
($A_{r}$ Fe =56; Zn = 65; Cl = 35,5 ; Sn = 119).
Betul$\begin{alignedat}{1}\mbox{massa Zn} & =1,95\mbox{ gram}\\
\mbox{mol Zn} & =\frac{\mbox{massa Zn}}{A_{r}\mbox{Zn}}\\
& =\frac{1,95}{65}\\
& =0,03\mbox{ mol}\\
\mbox{massa Fe} & =1,12\mbox{ gram}\\
\mbox{mol Fe} & =\frac{\mbox{massa Fe}}{A_{r}\mbox{ Fe}}\\
& =\frac{1,12}{56}\\
& =0,02\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
\mbox{FeCl}_{x}\\
\\
\mbox{mol Zn}\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{Zn}\\
& 0,03\mbox{ mol}\\
: & \mbox{mol Fe}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{ZnCl}_{2}\\
\\
= & 0,03\\
= & 3
\end{array}$ $\begin{array}{cc}
+ & \mbox{Fe}\\
& 0,02\mbox{ mol}\\
: & 0,02\\
: & 2
\end{array}$Perbandingan koefisien = perbandingan mol
$\mbox{FeCl}_{x}\mbox{+3Zn}$$\rightarrow\mbox{ZnCl}_{2}$$\mbox{+2Fe}$
Reaksi dikatakan setara jika jumlah atom di ruas kiri = jumlah atom
di ruas kanan, sehingga:Jumlah atom Fe di kanan = 2, sehingga koefisien $\mbox{FeCl}_{x}$
di kiri = 2.Jumlah atom Zn di kiri = 3, sehingga koefisien $\mbox{ZnCl}_{2}$
di kanan = 3$2\mbox{FeCl}_{x}+3\mbox{Zn}$$\rightarrow3\mbox{ZnCl}_{2}$$\mbox{+2Fe}$
Jumlah atom Cl di kiri = jumlah atom Cl di kanan
$\begin{alignedat}{1}2x & =3\times2\\
2x & =6\\
x & =3.
\end{alignedat}
$Salah$\begin{alignedat}{1}\mbox{massa Zn} & =1,95\mbox{ gram}\\
\mbox{mol Zn} & =\frac{\mbox{massa Zn}}{A_{r}\mbox{Zn}}\\
& =\frac{1,95}{65}\\
& =0,03\mbox{ mol}\\
\mbox{massa Fe} & =1,12\mbox{ gram}\\
\mbox{mol Fe} & =\frac{\mbox{massa Fe}}{A_{r}\mbox{ Fe}}\\
& =\frac{1,12}{56}\\
& =0,02\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
\mbox{FeCl}_{x}\\
\\
\mbox{mol Zn}\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{Zn}\\
& 0,03\mbox{ mol}\\
: & \mbox{mol Fe}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{ZnCl}_{2}\\
\\
= & 0,03\\
= & 3
\end{array}$ $\begin{array}{cc}
+ & \mbox{Fe}\\
& 0,02\mbox{ mol}\\
: & 0,02\\
: & 2
\end{array}$Perbandingan koefisien = perbandingan mol
$\mbox{FeCl}_{x}\mbox{+3Zn}$$\rightarrow\mbox{ZnCl}_{2}$$\mbox{+2Fe}$
Reaksi dikatakan setara jika jumlah atom di ruas kiri = jumlah atom
di ruas kanan, sehingga:Jumlah atom Fe di kanan = 2, sehingga koefisien $\mbox{FeCl}_{x}$
di kiri = 2.Jumlah atom Zn di kiri = 3, sehingga koefisien $\mbox{ZnCl}_{2}$
di kanan = 3$2\mbox{FeCl}_{x}+3\mbox{Zn}$$\rightarrow3\mbox{ZnCl}_{2}$$\mbox{+2Fe}$
Jumlah atom Cl di kiri = jumlah atom Cl di kanan
$\begin{alignedat}{1}2x & =3\times2\\
2x & =6\\
x & =3.
\end{alignedat}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
29,6 gram padatan $\mbox{X}\left(\mbox{NO}_{3}\right)_{2}$ mengalami penguraian membentuk oksida logam, nitrogen dioksida dan oksigen. Jika massa gas yang dihasilkan 21,6 gram, tentukan nama logam X tersebut !
Betul$\begin{alignedat}{1}\mbox{m }\mbox{X}\left(\mbox{NO}_{3}\right)_{2} & =29,6\mbox{ gram}\\
\mbox{m O}_{2}+\mbox{m NO}_{2} & =21,6\mbox{ gram}\\
\mbox{m oksida logam} & =29,6-21,6\mbox{ gram}\\
& =8\mbox{ gram}\\
\mbox{n }\mbox{X}\left(\mbox{NO}_{3}\right)_{2} & =\frac{29,6}{A_{r}\mbox{ X}+124}\mbox{mol}\\
\mbox{n XO} & =\frac{8}{A_{r}\mbox{ X}+16}\mbox{ mol}
\end{alignedat}
$Ratio mol $\mbox{X}\left(\mbox{NO}_{3}\right)_{2}=$mol $\mbox{XO}$ (koefisien reaksi sama)
$\begin{alignedat}{1}\frac{29,6}{A_{r}\mbox{ X}+124} & =\frac{8}{A_{r}\mbox{ X}+16}\\
8A_{r}\mbox{ X}+992 & =29,6A_{r}\mbox{ X}+473,6\\
21,6A_{r}\mbox{X} & =518,4\\
A_{r}\mbox{ X} & =\frac{518,4}{21,6}\\
& =24
\end{alignedat}
$Logam X dengan nilai $A_{r}$ = 24 yaitu logam Mg.
Salah$\begin{alignedat}{1}\mbox{m }\mbox{X}\left(\mbox{NO}_{3}\right)_{2} & =29,6\mbox{ gram}\\
\mbox{m O}_{2}+\mbox{m NO}_{2} & =21,6\mbox{ gram}\\
\mbox{m oksida logam} & =29,6-21,6\mbox{ gram}\\
& =8\mbox{ gram}\\
\mbox{n }\mbox{X}\left(\mbox{NO}_{3}\right)_{2} & =\frac{29,6}{A_{r}\mbox{ X}+124}\mbox{mol}\\
\mbox{n XO} & =\frac{8}{A_{r}\mbox{ X}+16}\mbox{ mol}
\end{alignedat}
$Ratio mol $\mbox{X}\left(\mbox{NO}_{3}\right)_{2}=$mol $\mbox{XO}$ (koefisien reaksi sama)
$\begin{alignedat}{1}\frac{29,6}{A_{r}\mbox{ X}+124} & =\frac{8}{A_{r}\mbox{ X}+16}\\
8A_{r}\mbox{ X}+992 & =29,6A_{r}\mbox{ X}+473,6\\
21,6A_{r}\mbox{X} & =518,4\\
A_{r}\mbox{ X} & =\frac{518,4}{21,6}\\
& =24
\end{alignedat}
$Logam X dengan nilai $A_{r}$ = 24 yaitu logam Mg.
-
Pertanyaan ke 3 dari 5
3. Pertanyaan
Karbon monoksida merupakan reduktor pada proses pengolahan logam besi pada proses tanur. Pembentukan gas karbon monoksida diperoleh dari reaksi berikut:
$\mbox{C}+\mbox{CO}_{2}\rightarrow\mbox{CO}_{2}$
$\mbox{CO}_{2}+\mbox{C}\rightarrow2\mbox{CO}$
Massa maksimum karbon monoksida yang terbentuk dari 60 gram karbon adalah … .
Betul$\begin{alignedat}{1}m_{\mbox{karbon}} & =60\mbox{ g}\\
n_{\mbox{karbon}} & =\frac{60}{12}\mbox{ mol}\\
& =5\mbox{mol}
\end{alignedat}
$Karbon digunakan sebagai reaktan di kedua reaksi, sehingga jumlah karbon dibagi 2.
$\begin{array}{cccccc}
& \mbox{C} & + & \mbox{O}_{2} & & \mbox{CO}_{2}\\
& 2,5\mbox{ mol} & & 2,5\mbox{ mol} & & 2,5\mbox{ mol}\\
& \mbox{CO}_{2} & + & \mbox{C} & \rightarrow & 2\mbox{CO}\\
& 2,5\mbox{ mol} & & 2,5\mbox{ mol} & & 5\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{CO}} & =5\mbox{ mol}\\
m_{CO} & =n\times M_{r}\mbox{ CO}\\
& =5\times28\mbox{ g}\\
& =140\mbox{ g}.
\end{alignedat}
$Salah$\begin{alignedat}{1}m_{\mbox{karbon}} & =60\mbox{ g}\\
n_{\mbox{karbon}} & =\frac{60}{12}\mbox{ mol}\\
& =5\mbox{mol}
\end{alignedat}
$Karbon digunakan sebagai reaktan di kedua reaksi, sehingga jumlah karbon dibagi 2.
$\begin{array}{cccccc}
& \mbox{C} & + & \mbox{O}_{2} & & \mbox{CO}_{2}\\
& 2,5\mbox{ mol} & & 2,5\mbox{ mol} & & 2,5\mbox{ mol}\\
& \mbox{CO}_{2} & + & \mbox{C} & \rightarrow & 2\mbox{CO}\\
& 2,5\mbox{ mol} & & 2,5\mbox{ mol} & & 5\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{CO}} & =5\mbox{ mol}\\
m_{CO} & =n\times M_{r}\mbox{ CO}\\
& =5\times28\mbox{ g}\\
& =140\mbox{ g}.
\end{alignedat}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
Satu gram sampel $\mbox{MgCO}_{3}$ direaksikan ke dalam $50\mbox{ cm}^{3}$ larutan $\mbox{HCl}$ 0,5 M.
$\mbox{MgCO}_{3}(s)\mbox{+2HCl}(aq)$ $\rightarrow\mbox{MgCl}(aq)$ $+\mbox{CO}_{2}(g)$ $+\mbox{H}_{2}\mbox{O}(\ell)$
Dibutuhkan $18\mbox{ cm}^{3}$ larutan NaOH 0,5 M untuk menetralkan larutan tersebut. Tentukan persen kemurnian $\mbox{MgCO}_{3}$ terhadap massa sampelnya.
Betul$\begin{alignedat}{1}n\mbox{ HCl awal} & =M\times v\\
& =0,5\times0,05\mbox{ mol}\\
& =0,025\mbox{ mol}\\
n\mbox{ NaOH} & =M\times v\\
& =0,5\times0,018\mbox{ mol}\\
& =0,009\mbox{ mol}\\
n\mbox{ NaOH} & =\mbox{n HCl\ reaksi}\\
& =0,009\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
\mbox{HCl}(aq)\\
0,009\mbox{ mol}\\
\mbox{MgCO}_{3}(s)
\end{array}$ $\begin{array}{cc}
+ & \mbox{NaOH}(aq)\\
& 0,009\mbox{ mol}\\
+ & 2\mbox{HCl}(aq)
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{NaCl}(aq)\\
\\
\rightarrow & \mbox{MgCl}_{2}(aq)
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}\mbox{O}(\ell)\\
\\
+ & \mbox{CO}_{2}(g)
\end{array}$ $\begin{array}{cc}
\\
\\
+ & \mbox{H}_{2}\mbox{O}(\ell)
\end{array}$$\begin{aligned}n\mbox{ HCl sisa} & =n\mbox{ HCl awal}-n\mbox{ HCl reaksi}\\
0,009\mbox{ mol} & =0,025\mbox{ mol}-n\mbox{ HCl reaksi}\\
n\mbox{ HCl reaksi} & =0,016\mbox{ mol}
\end{aligned}
$$\begin{aligned}n\mbox{ MgCO}_{3} & =\frac{1}{2}\times n\mbox{ HCl}\\
& =\frac{1}{2}\times0,016\mbox{ mol}\\
& =0,008\mbox{ mol}
\end{aligned}
$$\begin{array}{c}
\mbox{MgCO}_{3}(s)\\
\frac{1}{2}\times0,016\mbox{ mol}\\
=0,008\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{HCl}(aq)\\
& 0,016\mbox{ mol}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{MgCl}_{2}(aq)\\
\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{CO}_{2}(g)\\
\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}\mbox{O}(\ell)\\
\\
\\
\end{array}$$\begin{alignedat}{1}\mbox{m MgCO}_{3} & =0,008\times84\mbox{ gram}\\
& =0,672\mbox{ gram}\\
\%\mbox{ kemurnian} & =\frac{m\mbox{ MgCO}_{3}}{m\mbox{ sampel}}\times100\%\\
& =\frac{0,672}{1}\times100\%\\
& =67,2\%.
\end{alignedat}
$Salah$\begin{alignedat}{1}n\mbox{ HCl awal} & =M\times v\\
& =0,5\times0,05\mbox{ mol}\\
& =0,025\mbox{ mol}\\
n\mbox{ NaOH} & =M\times v\\
& =0,5\times0,018\mbox{ mol}\\
& =0,009\mbox{ mol}\\
n\mbox{ NaOH} & =\mbox{n HCl\ reaksi}\\
& =0,009\mbox{ mol}
\end{alignedat}
$$\begin{array}{c}
\mbox{HCl}(aq)\\
0,009\mbox{ mol}\\
\mbox{MgCO}_{3}(s)
\end{array}$ $\begin{array}{cc}
+ & \mbox{NaOH}(aq)\\
& 0,009\mbox{ mol}\\
+ & 2\mbox{HCl}(aq)
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{NaCl}(aq)\\
\\
\rightarrow & \mbox{MgCl}_{2}(aq)
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}\mbox{O}(\ell)\\
\\
+ & \mbox{CO}_{2}(g)
\end{array}$ $\begin{array}{cc}
\\
\\
+ & \mbox{H}_{2}\mbox{O}(\ell)
\end{array}$$\begin{aligned}n\mbox{ HCl sisa} & =n\mbox{ HCl awal}-n\mbox{ HCl reaksi}\\
0,009\mbox{ mol} & =0,025\mbox{ mol}-n\mbox{ HCl reaksi}\\
n\mbox{ HCl reaksi} & =0,016\mbox{ mol}
\end{aligned}
$$\begin{aligned}n\mbox{ MgCO}_{3} & =\frac{1}{2}\times n\mbox{ HCl}\\
& =\frac{1}{2}\times0,016\mbox{ mol}\\
& =0,008\mbox{ mol}
\end{aligned}
$$\begin{array}{c}
\mbox{MgCO}_{3}(s)\\
\frac{1}{2}\times0,016\mbox{ mol}\\
=0,008\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{HCl}(aq)\\
& 0,016\mbox{ mol}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{MgCl}_{2}(aq)\\
\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{CO}_{2}(g)\\
\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}\mbox{O}(\ell)\\
\\
\\
\end{array}$$\begin{alignedat}{1}\mbox{m MgCO}_{3} & =0,008\times84\mbox{ gram}\\
& =0,672\mbox{ gram}\\
\%\mbox{ kemurnian} & =\frac{m\mbox{ MgCO}_{3}}{m\mbox{ sampel}}\times100\%\\
& =\frac{0,672}{1}\times100\%\\
& =67,2\%.
\end{alignedat}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
16,73 gram garam hidrat $\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O}$ dilarutkan ke dalam air sehingga volumenya 500 mL. 20 mL larutan tersebut direaksikan dengan 23,4 mL larutan $\mbox{HCl}$ 0,2 M. Jika diketahui $M_{r}$ garam hidrat tersebut 286, berapakah nilai $x$ ?
BetulReaksi yang terjadi:
$\begin{array}{c}
\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O}(s)\\
\mbox{Na}_{2}\mbox{CO}_{3}(aq)
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Na}_{2}\mbox{CO}_{3}(aq)\\
+ & \mbox{2HCl}(aq)
\end{array}$ $\begin{array}{cc}
+ & \mbox{xH}_{2}\mbox{O}(\ell)\\
\rightarrow & \mbox{2NaCl}(aq)
\end{array}$ $\begin{array}{cc}
\\
+ & \mbox{H}_{2}\mbox{O}(\ell)
\end{array}$$\begin{alignedat}{1}n\mbox{ HCl} & =0,2\times0,0234\mbox{ mol}\\
& =4,68\times10^{-3}\mbox{mol}\\
n\mbox{ Na}_{2}\mbox{CO}_{3} & =\frac{1}{2}\times\mbox{n HCl}\\
& =\frac{1}{2}\times4,68\times10^{-3}\mbox{ mol}\\
& =2,34\times10^{-3}\mbox{ mol}
\end{alignedat}
$$\begin{aligned}n\mbox{Na}_{2}\mbox{CO}_{3}\mbox{ dalam 500 mL} & =\frac{500\mbox{ mL}}{20\mbox{ mL}}\times2,34\times10^{-3}\mbox{ mol}\\
& =0,0585\mbox{ mol}
\end{aligned}
$$\begin{aligned}n\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O} & =n\mbox{ Na}_{2}\mbox{CO}_{3}\\
& =0,0585\mbox{ mol}
\end{aligned}
$$\begin{aligned}M_{r} & =\frac{\mbox{massa}}{\mbox{mol}}\\
& =\frac{16,73}{0,0585}\\
& =286
\end{aligned}
$$\begin{aligned}M_{r}\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O} & =286\\
(2\times23)+(1\times12)+(3\times16)+x((2\times1)+(1\times16)) & =286\\
46+12+48+18x & =286\\
106+18x & =286\\
18x & =180\\
x & =10
\end{aligned}
$SalahReaksi yang terjadi:
$\begin{array}{c}
\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O}(s)\\
\mbox{Na}_{2}\mbox{CO}_{3}(aq)
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Na}_{2}\mbox{CO}_{3}(aq)\\
+ & \mbox{2HCl}(aq)
\end{array}$ $\begin{array}{cc}
+ & \mbox{xH}_{2}\mbox{O}(\ell)\\
\rightarrow & \mbox{2NaCl}(aq)
\end{array}$ $\begin{array}{cc}
\\
+ & \mbox{H}_{2}\mbox{O}(\ell)
\end{array}$$\begin{alignedat}{1}n\mbox{ HCl} & =0,2\times0,0234\mbox{ mol}\\
& =4,68\times10^{-3}\mbox{mol}\\
n\mbox{ Na}_{2}\mbox{CO}_{3} & =\frac{1}{2}\times\mbox{n HCl}\\
& =\frac{1}{2}\times4,68\times10^{-3}\mbox{ mol}\\
& =2,34\times10^{-3}\mbox{ mol}
\end{alignedat}
$$\begin{aligned}n\mbox{Na}_{2}\mbox{CO}_{3}\mbox{ dalam 500 mL} & =\frac{500\mbox{ mL}}{20\mbox{ mL}}\times2,34\times10^{-3}\mbox{ mol}\\
& =0,0585\mbox{ mol}
\end{aligned}
$$\begin{aligned}n\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O} & =n\mbox{ Na}_{2}\mbox{CO}_{3}\\
& =0,0585\mbox{ mol}
\end{aligned}
$$\begin{aligned}M_{r} & =\frac{\mbox{massa}}{\mbox{mol}}\\
& =\frac{16,73}{0,0585}\\
& =286
\end{aligned}
$$\begin{aligned}M_{r}\mbox{Na}_{2}\mbox{CO}_{3}\cdot x\mbox{H}_{2}\mbox{O} & =286\\
(2\times23)+(1\times12)+(3\times16)+x((2\times1)+(1\times16)) & =286\\
46+12+48+18x & =286\\
106+18x & =286\\
18x & =180\\
x & =10
\end{aligned}
$