Selamat Datang dan Selamat Belajar di Wardaya College!
Di sini, kamu akan belajar tentang Persentase Hasil melalui latihan soal interaktif dalam 3 tingkat kesulitan (mudah, sedang, sukar).
Apabila materi ini berguna, bagikan ke teman atau rekan kamu supaya mereka juga mendapatkan manfaatnya.
Latihan Soal Persentase Hasil (Mudah)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Pada suatu reaksi 1 mol hidrogen dengan oksigen menghasilkan 15 gram air. Berapakah persentase hasil reaksi tersebut?
($A_{r}$ H = 1; O = 16)
$2\mbox{H}_{2}+\mbox{O}_{2}$$\rightarrow2\mbox{H}_{2}\mbox{O}$
Betul$\begin{array}{ccccc}
2\mbox{H}_{2} & + & \mbox{O}_{2} & \rightarrow & 2\mbox{H}_{2}\mbox{O}\\
2 & : & 1 & : & 2
\end{array}$mol $\mbox{H}_{2}=1$ mol
$\begin{aligned}M_{r}\mbox{ H}_{2}\mbox{O} & =\left(2\times A_{r}\mbox{ H}\right)+\left(1\times A_{r}\mbox{ O}\right)\\
& =(2\times1)+(1\times16)\\
& =2+16\\
& =18
\end{aligned}
$$\begin{aligned}\mbox{mol H}_{2}\mbox{O} & =\frac{2}{2}\times\mbox{mol H}_{2}\\
& =\frac{2}{2}\times1\mbox{ mol}\\
& =1\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{massa H}_{2}\mbox{O}\mbox{teori} & =\mbox{mol H}_{2}\mbox{O}\times M_{r}\mbox{ H}_{2}\mbox{O}\\
& =1\times18\mbox{ g}\\
& =18\mbox{ g}
\end{aligned}
$$\begin{aligned}\mbox{persentase hasil} & =\frac{\mbox{massa sebenarnya}}{\mbox{massa teoritis}}\times100\%\\
& =\frac{15}{18}\times100\%\\
& =83,33\%.
\end{aligned}
$Salah$\begin{array}{ccccc}
2\mbox{H}_{2} & + & \mbox{O}_{2} & \rightarrow & 2\mbox{H}_{2}\mbox{O}\\
2 & : & 1 & : & 2
\end{array}$mol $\mbox{H}_{2}=1$ mol
$\begin{aligned}M_{r}\mbox{ H}_{2}\mbox{O} & =\left(2\times A_{r}\mbox{ H}\right)+\left(1\times A_{r}\mbox{ O}\right)\\
& =(2\times1)+(1\times16)\\
& =2+16\\
& =18
\end{aligned}
$$\begin{aligned}\mbox{mol H}_{2}\mbox{O} & =\frac{2}{2}\times\mbox{mol H}_{2}\\
& =\frac{2}{2}\times1\mbox{ mol}\\
& =1\mbox{ mol}
\end{aligned}
$$\begin{aligned}\mbox{massa H}_{2}\mbox{O}\mbox{teori} & =\mbox{mol H}_{2}\mbox{O}\times M_{r}\mbox{ H}_{2}\mbox{O}\\
& =1\times18\mbox{ g}\\
& =18\mbox{ g}
\end{aligned}
$$\begin{aligned}\mbox{persentase hasil} & =\frac{\mbox{massa sebenarnya}}{\mbox{massa teoritis}}\times100\%\\
& =\frac{15}{18}\times100\%\\
& =83,33\%.
\end{aligned}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
Pada proses penguraian 10 gram $\mbox{CaCO}_{3}$ menghasilkan 4 gram CaO berdasarkan reaksi berikut
$\mbox{CaCO}_{3}(s)$$\rightarrow\mbox{CaO}(s)$$+\mbox{CO}_{2}(g)$
Tentukan persentase hasil reaksi tersebut?
($A_{r}$ CA = 40, C = 12, O = 16).
Betul$\begin{alignedat}{1}n_{\mbox{CaCO}_{3}} & =\frac{10}{100}\mbox{ mol}\\
& =0,1\mbox{ mol}\\
n_{\mbox{CaO}} & =\frac{1}{1}\times0,1\mbox{ mol}\\
& =0,1\mbox{ mol}\\
m_{\mbox{CaO}} & =0,1\mbox{ mol}\times(40+16)\mbox{g/mol}\\
& =5,6\mbox{ g}\\
\%\mbox{hasil} & =\frac{\mbox{m yg didapat}}{\mbox{m teori}}\times100\%\\
& =\frac{4}{5,6}\times100\%\\
& =71,4\%.
\end{alignedat}
$Salah$\begin{alignedat}{1}n_{\mbox{CaCO}_{3}} & =\frac{10}{100}\mbox{ mol}\\
& =0,1\mbox{ mol}\\
n_{\mbox{CaO}} & =\frac{1}{1}\times0,1\mbox{ mol}\\
& =0,1\mbox{ mol}\\
m_{\mbox{CaO}} & =0,1\mbox{ mol}\times(40+16)\mbox{g/mol}\\
& =5,6\mbox{ g}\\
\%\mbox{hasil} & =\frac{\mbox{m yg didapat}}{\mbox{m teori}}\times100\%\\
& =\frac{4}{5,6}\times100\%\\
& =71,4\%.
\end{alignedat}
$ -
Pertanyaan ke 3 dari 5
3. Pertanyaan
16 L $\mbox{N}_{2}\mbox{O}_{3}$ mengalami penguraian berdasarkan reaksi berikut.
$2\mbox{N}_{2}\mbox{O}_{3}(g)\rightarrow\mbox{N}_{2}\mbox{O}_{4}(g)+2\mbox{NO(g)}$
Jika persentase hasil reaksi tersebut 75%, tentukan volume gas $\mbox{N}_{2}\mbox{O}_{4}$ yang dihasilkan dari reaksi tersebut.
Betul$\begin{array}{c}
2\mbox{N}_{2}\mbox{O}_{3}(g)\\
2\\
16\mbox{ L}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{N}_{2}\mbox{O}_{4}(g)\\
: & 1\\
& \frac{1}{2}\times16\mbox{ L}\\
& =8\mbox{ L}
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{NO(g)}\\
: & 2\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{alignedat}{1}V_{\mbox{teori}} & =8\mbox{ L}\\
\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
75\% & =\frac{V_{\mbox{aktual}}}{8\mbox{ L}}\times100\%\\
V_{\mbox{aktual}} & =6\mbox{ L}.
\end{alignedat}
$Salah$\begin{array}{c}
2\mbox{N}_{2}\mbox{O}_{3}(g)\\
2\\
16\mbox{ L}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{N}_{2}\mbox{O}_{4}(g)\\
: & 1\\
& \frac{1}{2}\times16\mbox{ L}\\
& =8\mbox{ L}
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{NO(g)}\\
: & 2\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{alignedat}{1}V_{\mbox{teori}} & =8\mbox{ L}\\
\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
75\% & =\frac{V_{\mbox{aktual}}}{8\mbox{ L}}\times100\%\\
V_{\mbox{aktual}} & =6\mbox{ L}.
\end{alignedat}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
1,2 L gas oksigen bereaksi dengan gas sulfur dioksida$\left(\mbox{NO}_{2}\right)$ berlebih menghasilkan 2 L sulfur trioksida $\left(\mbox{SO}_{3}\right)$.
$2\mbox{SO}_{2}\mbox{+O}_{2}$$\rightarrow2\mbox{SO}_{3}$
Berapakah persentase hasil reaksi tersebut ?
Betul$\begin{array}{ccccc}
2\mbox{SO}_{2} & + & \mbox{O}_{2} & \rightarrow & 2\mbox{SO}_{3}\\
& & 1,2\mbox{ L} & & \frac{2}{1}\times1,2\mbox{ L}\\
& & & & =2,4\mbox{ L}
\end{array}$$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{2}{2,4}\times100\%\\
& =83,3\%.
\end{alignedat}
$Salah$\begin{array}{ccccc}
2\mbox{SO}_{2} & + & \mbox{O}_{2} & \rightarrow & 2\mbox{SO}_{3}\\
& & 1,2\mbox{ L} & & \frac{2}{1}\times1,2\mbox{ L}\\
& & & & =2,4\mbox{ L}
\end{array}$$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{2}{2,4}\times100\%\\
& =83,3\%.
\end{alignedat}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Perhatikan reaksi berikut.
$2\mbox{NO}(g)+\mbox{O}_{2}(g)$$\rightarrow2\mbox{NO}_{2}(g)$
Dalam reaksi di atas gas NO tepat bereaksi dengan gas oksigen dan dihasilkan 4 L gas $\mbox{NO}_{2}$. Jika persentase hasil reaksi tersebut 80%, berapakah volume gas yang hilang ?
Betul$\begin{alignedat}{1}V_{\mbox{aktual}} & =4\mbox{ L}\\
\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
80\% & =\frac{4\mbox{ L}}{V_{\mbox{teori}}}\times100\%\\
V_{\mbox{teori}} & =5\mbox{ L}\\
V_{\mbox{hilang}} & =V_{\mbox{teori}}-V_{\mbox{aktual}}\\
& =5-4\mbox{ L}\\
& =1\mbox{ L.}
\end{alignedat}
$Salah$\begin{alignedat}{1}V_{\mbox{aktual}} & =4\mbox{ L}\\
\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
80\% & =\frac{4\mbox{ L}}{V_{\mbox{teori}}}\times100\%\\
V_{\mbox{teori}} & =5\mbox{ L}\\
V_{\mbox{hilang}} & =V_{\mbox{teori}}-V_{\mbox{aktual}}\\
& =5-4\mbox{ L}\\
& =1\mbox{ L.}
\end{alignedat}
$
Latihan Soal Persentase Hasil (Sedang)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Dekomposisi gas $\mbox{SO}_{3}$ menghasilkan gas $\mbox{SO}_{2}$ dan $\mbox{O}_{2}$. 5 L $\mbox{SO}_{3}$ menghasilkan 2 L gas $\mbox{O}_{2}$. Tentukan volume gas $\mbox{SO}_{2}$ yang dihasilkan pada reaksi tersebut.
Betul$\begin{array}{c}
\mbox{2SO}_{3}(g)\\
5\mbox{ L}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & 2\mbox{SO}_{2}(g)\\
& \frac{2}{2}\times5\mbox{ L}\\
& 5\mbox{ L}
\end{array}$ $\begin{array}{cc}
+ & \mbox{O}_{2}(g)\\
& \frac{1}{2}\times5\mbox{ L}\\
& =2,5\mbox{ L}
\end{array}$$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{2}{2,5}\times100\%\\
& =80\%\\
V_{\mbox{aktual SO}_{2}} & =80\%\times5\mbox{ L}\\
& =4\mbox{ L.}
\end{alignedat}
$Salah$\begin{array}{c}
\mbox{2SO}_{3}(g)\\
5\mbox{ L}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & 2\mbox{SO}_{2}(g)\\
& \frac{2}{2}\times5\mbox{ L}\\
& 5\mbox{ L}
\end{array}$ $\begin{array}{cc}
+ & \mbox{O}_{2}(g)\\
& \frac{1}{2}\times5\mbox{ L}\\
& =2,5\mbox{ L}
\end{array}$$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{2}{2,5}\times100\%\\
& =80\%\\
V_{\mbox{aktual SO}_{2}} & =80\%\times5\mbox{ L}\\
& =4\mbox{ L.}
\end{alignedat}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
2,6 gram asetilen $\left(\mbox{C}_{2}\mbox{H}_{2}\right)$ dapat dihasilkan dari reaksi kalsium karbida dengan air, berdasarkan reaksi berikut.
$\mbox{CaC}_{2}+2\mbox{H}_{2}\mbox{O}$$\rightarrow\mbox{Ca(OH)}_{2}+\mbox{C}_{2}\mbox{H}_{2}$
Berapakah massa kalsium karbida yang direaksikan jika persentase hasil reaksi tersebut 80% ?
Betul$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\
80\% & =\frac{2,6\mbox{ gram}}{m_{\mbox{teori}}}\times100\%\\
m_{\mbox{teori}} & =3,25\mbox{ gram}
\end{alignedat}
$massa teori $\mbox{C}_{2}\mbox{H}_{2}$= 3,25 gram.
mol teori $\mbox{C}_{2}\mbox{H}_{2}$$=\frac{3,25}{(24+2)}$$=0,125$
mol.$\begin{array}{c}
\mbox{CaC}_{2}\\
\frac{1}{1}\times0,125\mbox{ mol}\\
=0,125\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Ca(OH)}_{2}\\
\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{C}_{2}\mbox{H}_{2}\\
& 0,125\mbox{ mol}\\
\\
\end{array}$$\begin{aligned}\mbox{m CaC}_{2} & =0,125\mbox{ mol}\times(40+24)\mbox{g/mol}\\
& =8\mbox{ gram}.
\end{aligned}
$Salah$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\
80\% & =\frac{2,6\mbox{ gram}}{m_{\mbox{teori}}}\times100\%\\
m_{\mbox{teori}} & =3,25\mbox{ gram}
\end{alignedat}
$massa teori $\mbox{C}_{2}\mbox{H}_{2}$= 3,25 gram.
mol teori $\mbox{C}_{2}\mbox{H}_{2}$$=\frac{3,25}{(24+2)}$$=0,125$
mol.$\begin{array}{c}
\mbox{CaC}_{2}\\
\frac{1}{1}\times0,125\mbox{ mol}\\
=0,125\mbox{ mol}
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{Ca(OH)}_{2}\\
\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{C}_{2}\mbox{H}_{2}\\
& 0,125\mbox{ mol}\\
\\
\end{array}$$\begin{aligned}\mbox{m CaC}_{2} & =0,125\mbox{ mol}\times(40+24)\mbox{g/mol}\\
& =8\mbox{ gram}.
\end{aligned}
$ -
Pertanyaan ke 3 dari 5
3. Pertanyaan
Gas metana $\left(\mbox{CH}_{4}\right)$ dapat dihasilkan dari reaksi karbon monoksida (CO) dengan gas hidrogen.
$\mbox{CO}(g)+3\mbox{H}_{2}(g)$$\rightarrow\mbox{CH}_{4}(g)$ $+\mbox{H}_{2}\mbox{O}(g)$
Dari reaksi tersebut terdapat selisih 2 L gas $\mbox{CH}_{4}$ dari yang seharusnya dihasilkan. Jika persentase hasil reaksi 75%, berapakah volume aktual yang diperoleh ?
Betul$\begin{aligned}V_{\mbox{teori}}-V_{\mbox{aktual}} & =2\mbox{ L}\\
V_{\mbox{teori}} & =V_{\mbox{aktual}}+2\mbox{ L}
\end{aligned}
$$\begin{aligned}\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
75\% & =\frac{V_{\mbox{aktual}}}{V_{\mbox{aktual}}+2\mbox{ L}}\times100\%\\
75\%(V_{\mbox{aktual}}+2\mbox{ L}) & =V_{\mbox{aktual}}\times100\%\\
V_{\mbox{aktual}} & =6\mbox{ L}
\end{aligned}
$Salah$\begin{aligned}V_{\mbox{teori}}-V_{\mbox{aktual}} & =2\mbox{ L}\\
V_{\mbox{teori}} & =V_{\mbox{aktual}}+2\mbox{ L}
\end{aligned}
$$\begin{aligned}\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
75\% & =\frac{V_{\mbox{aktual}}}{V_{\mbox{aktual}}+2\mbox{ L}}\times100\%\\
75\%(V_{\mbox{aktual}}+2\mbox{ L}) & =V_{\mbox{aktual}}\times100\%\\
V_{\mbox{aktual}} & =6\mbox{ L}
\end{aligned}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
5,75 gram logam natrium terdapat dalam 15,5 gram campuran logam natrium-kalium. Campuran logam ini dimasukkan ke dalam 500 mL air dan terbentuk 4 L gas hidrogen. Berapakah percentage yield reaksi? (Ar Na = 23 , K = 39, O = 16, H = 1)
Betul$\begin{array}{c}
2\mbox{Na}(s)\\
5,75\mbox{ gram}\\
\frac{5,75\mbox{ gram}}{23}\\
=0,25\mbox{ mol}\\
\\
2\mbox{K}(s)\\
(15,5-5,75)\mbox{ gram}\\
\frac{9,75}{39}\mbox{ mol}\\
0,25\mbox{ mol}\\
\\
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}(\ell)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
+ & 2\mbox{H}_{2}\mbox{O}(\ell)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & 2\mbox{NaOH}(aq)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\rightarrow & 2\mbox{KOH}(aq)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}(g)\\
\\
\\
& \frac{1}{2}\times0,25\mbox{ mol}\\
& 0,125\mbox{ mol}\\
+ & \mbox{H}_{2}(g)\\
\\
\\
\,^{\,^{\,}} & \frac{1}{2}\times0,25\mbox{ mol}\\
& 0,125\mbox{ mol}
\end{array}$$\begin{aligned}n_{\mbox{total H}_{2}} & =0,125\mbox{ mol}+0,125\mbox{ mol}\\
& =0,25\mbox{ mol}
\end{aligned}
$$\begin{alignedat}{1}V_{\mbox{total H}_{2}} & =n\times22,4\mbox{ L}\\
& =0,25\times22,4\mbox{ L}\\
& =5,6\mbox{ L}\\
\%\mbox{hasil} & =\frac{4}{5,6}\times100\%\\
& =71,4\%.
\end{alignedat}
$Salah$\begin{array}{c}
2\mbox{Na}(s)\\
5,75\mbox{ gram}\\
\frac{5,75\mbox{ gram}}{23}\\
=0,25\mbox{ mol}\\
\\
2\mbox{K}(s)\\
(15,5-5,75)\mbox{ gram}\\
\frac{9,75}{39}\mbox{ mol}\\
0,25\mbox{ mol}\\
\\
\end{array}$ $\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}(\ell)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
+ & 2\mbox{H}_{2}\mbox{O}(\ell)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\\
\end{array}$ $\begin{array}{cc}
\rightarrow & 2\mbox{NaOH}(aq)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\rightarrow & 2\mbox{KOH}(aq)\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\\
\end{array}$ $\begin{array}{cc}
+ & \mbox{H}_{2}(g)\\
\\
\\
& \frac{1}{2}\times0,25\mbox{ mol}\\
& 0,125\mbox{ mol}\\
+ & \mbox{H}_{2}(g)\\
\\
\\
\,^{\,^{\,}} & \frac{1}{2}\times0,25\mbox{ mol}\\
& 0,125\mbox{ mol}
\end{array}$$\begin{aligned}n_{\mbox{total H}_{2}} & =0,125\mbox{ mol}+0,125\mbox{ mol}\\
& =0,25\mbox{ mol}
\end{aligned}
$$\begin{alignedat}{1}V_{\mbox{total H}_{2}} & =n\times22,4\mbox{ L}\\
& =0,25\times22,4\mbox{ L}\\
& =5,6\mbox{ L}\\
\%\mbox{hasil} & =\frac{4}{5,6}\times100\%\\
& =71,4\%.
\end{alignedat}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Seorang siswa melakukan eksperimen pembakaran gas metana $\left(\mbox{CH}_{4}\right)$ dan etana $\left(\mbox{C}_{2}\mbox{H}_{6}\right)$ dengan oksigen berlebih.
1) Percobaan I: Pembakaran 0,2 mol gas metana menghasilkan 7,8 gram gas $\mbox{CO}_{2}$.
2) Percobaan II: pembakaran 0,2 mol gas etana dihasilkan 15,4 gram gas $\mbox{CO}_{2}$.
Pernyataan berikut ini yang benar mengenai eksperimen ini adalah … .
Betul$\begin{array}{cc}
\mbox{CH}_{4} & +2\mbox{O}_{2}\\
0,2\mbox{ mol}\\
\\
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{CO}_{2}\\
& \frac{1}{1}\times0,2\mbox{ mol}\\
& =0,2\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\end{array}$$\begin{alignedat}{1}m_{\mbox{CO}_{2}} & =0,2\times44\mbox{ gram}\\
& =8,8\mbox{ gram}\\
\%\mbox{hasil} & =\frac{7,8}{8,8}\times100\%\\
& =88,6\%
\end{alignedat}
$$\begin{array}{cc}
\mbox{C}_{2}\mbox{H}_{4} & +\mbox{3O}_{2}\\
0,2\mbox{ mol}\\
\\
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CO}_{2}\\
& \frac{2}{1}\times0,2\mbox{ mol}\\
& =0,4\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\end{array}$$\begin{alignedat}{1}m_{\mbox{CO}_{2}} & =0,4\times44\mbox{ gram}\\
& =17,6\mbox{ gram}\\
\%\mbox{hasil} & =\frac{15,4}{17,6}\times100\%\\
& =87,5\%.
\end{alignedat}
$Salah$\begin{array}{cc}
\mbox{CH}_{4} & +2\mbox{O}_{2}\\
0,2\mbox{ mol}\\
\\
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{CO}_{2}\\
& \frac{1}{1}\times0,2\mbox{ mol}\\
& =0,2\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\end{array}$$\begin{alignedat}{1}m_{\mbox{CO}_{2}} & =0,2\times44\mbox{ gram}\\
& =8,8\mbox{ gram}\\
\%\mbox{hasil} & =\frac{7,8}{8,8}\times100\%\\
& =88,6\%
\end{alignedat}
$$\begin{array}{cc}
\mbox{C}_{2}\mbox{H}_{4} & +\mbox{3O}_{2}\\
0,2\mbox{ mol}\\
\\
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CO}_{2}\\
& \frac{2}{1}\times0,2\mbox{ mol}\\
& =0,4\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\end{array}$$\begin{alignedat}{1}m_{\mbox{CO}_{2}} & =0,4\times44\mbox{ gram}\\
& =17,6\mbox{ gram}\\
\%\mbox{hasil} & =\frac{15,4}{17,6}\times100\%\\
& =87,5\%.
\end{alignedat}
$
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Gas hidrogen dan nitrogen dapat diperoleh dari reaksi dekomposisi amonia $\left(\mbox{NH}_{3}\right)$ berdasarkan reaksi berikut.
$2\mbox{NH}_{3}(g)\rightarrow\mbox{3H}_{2}+\mbox{N}_{2}(g)$
Jika 3,4 gram gas amonia menghasilkan 2,1 gram gas $\mbox{N}_{2}$, berapakah volume gas $\mbox{H}_{2}$ yang dihasilkan dari reaksi tersebut pada kondisi RTP.
Betul$\begin{array}{c}
\mbox{2NH}_{3}(g)\\
3,4\mbox{ gram}\\
n=\frac{3,4}{17}\mbox{mol}\\
=0,2\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{3H}_{2}(g)\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$ $\begin{array}{cc}
+ & \mbox{N}_{2}(g)\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{alignedat}{1}n_{\mbox{N}_{2}} & =\frac{1}{2}\times0,2\mbox{ mol}\\
& =0,1\mbox{ mol}\\
m_{\mbox{N}_{2}} & =0,1\times28\mbox{ gram}\\
& =2,8\mbox{ gram}\\
\%\mbox{hasil} & =\frac{2,1}{2,8}\times100\%\\
& =75\%\\
n_{\mbox{H}_{2}} & =\frac{3}{2}\times0,2\mbox{ mol}\\
& =0,3\mbox{ mol}\\
V_{\mbox{teori H}_{2}} & =0,3\times24\mbox{ dm}^{3}\\
& =7,2\mbox{ dm}^{3}\\
V_{\mbox{aktual H}_{2}} & =75\%\times7,2\mbox{ dm}^{3}\\
& =5,4\mbox{ dm}^{3}.
\end{alignedat}
$Salah$\begin{array}{c}
\mbox{2NH}_{3}(g)\\
3,4\mbox{ gram}\\
n=\frac{3,4}{17}\mbox{mol}\\
=0,2\mbox{ mol}
\end{array}$ $\begin{array}{cc}
\rightarrow & \mbox{3H}_{2}(g)\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$ $\begin{array}{cc}
+ & \mbox{N}_{2}(g)\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{alignedat}{1}n_{\mbox{N}_{2}} & =\frac{1}{2}\times0,2\mbox{ mol}\\
& =0,1\mbox{ mol}\\
m_{\mbox{N}_{2}} & =0,1\times28\mbox{ gram}\\
& =2,8\mbox{ gram}\\
\%\mbox{hasil} & =\frac{2,1}{2,8}\times100\%\\
& =75\%\\
n_{\mbox{H}_{2}} & =\frac{3}{2}\times0,2\mbox{ mol}\\
& =0,3\mbox{ mol}\\
V_{\mbox{teori H}_{2}} & =0,3\times24\mbox{ dm}^{3}\\
& =7,2\mbox{ dm}^{3}\\
V_{\mbox{aktual H}_{2}} & =75\%\times7,2\mbox{ dm}^{3}\\
& =5,4\mbox{ dm}^{3}.
\end{alignedat}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
Perhatikan reaksi berikut.
$\mbox{2NH}_{3}(g)$$+\mbox{NaClO}(s)$$\rightarrow\mbox{N}_{2}\mbox{H}_{4}(g)$$+\mbox{NaCl}(s)$$+\mbox{H}_{2}\mbox{O}(\ell)$
51 gram gas amonia bereaksi dengan $\mbox{NaClO}$ dihasilkan produk sesuai reaksi di atas. Dalam reaksi tersebut, terdapat 6 gram gas $\mbox{N}_{2}\mbox{H}_{4}$ yang hilang. Tentukan persentase hasil reaksi tersebut.
Betul$\begin{alignedat}{1}m_{\mbox{teori}}-m_{\mbox{aktual}} & =6\mbox{ gram}\\
m_{\mbox{teori}} & =6+m_{\mbox{aktual}}\\
n_{\mbox{teori}} & =\frac{6+m_{\mbox{aktual}}}{32}\\
n_{\mbox{NH}_{3}} & =\frac{2}{1}\times\frac{6+m_{\mbox{aktual}}}{32}\\
& =\frac{6+m_{\mbox{aktual}}}{16}\\
m_{\mbox{NH}_{3}} & =n\times M_{r}\\
51\mbox{ gram} & =\frac{6+m_{\mbox{aktual}}}{16}\times17\\
m_{\mbox{aktual}} & =48\mbox{ gram}\\
\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\
& =\frac{48}{48+6}\times100\%\\
& =88,9\%.
\end{alignedat}
$Salah$\begin{alignedat}{1}m_{\mbox{teori}}-m_{\mbox{aktual}} & =6\mbox{ gram}\\
m_{\mbox{teori}} & =6+m_{\mbox{aktual}}\\
n_{\mbox{teori}} & =\frac{6+m_{\mbox{aktual}}}{32}\\
n_{\mbox{NH}_{3}} & =\frac{2}{1}\times\frac{6+m_{\mbox{aktual}}}{32}\\
& =\frac{6+m_{\mbox{aktual}}}{16}\\
m_{\mbox{NH}_{3}} & =n\times M_{r}\\
51\mbox{ gram} & =\frac{6+m_{\mbox{aktual}}}{16}\times17\\
m_{\mbox{aktual}} & =48\mbox{ gram}\\
\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\
& =\frac{48}{48+6}\times100\%\\
& =88,9\%.
\end{alignedat}
$ -
Pertanyaan ke 3 dari 5
3. Pertanyaan
5 gram amalgam (campuran logam) yang terdiri atas logam Zn dan Cu dimasukkan ke dalam larutan asam klorida 500 mL. Dari reaksi tersebut dihasilkan 800 $\mbox{cm}^{3}$ gas hidrogen (RTP). Jika persentase hasil reaksi 83,3%, maka massa Zn dan Cu berturut-turut adalah … .
Betul$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
88,3\% & =\frac{800}{V_{\mbox{teori}}}\times100\%\\
V_{\mbox{teori}} & =960\mbox{ cm}^{3}\\
n_{\mbox{H}_{2}} & =\frac{960}{24000}\mbox{mol}\\
& =0,04\mbox{ mol}\\
\\
\end{alignedat}
$$\begin{array}{cc}
\mbox{Zn} & +2\mbox{HCl}\\
\frac{1}{1}\times0,04\mbox{ mol}\\
=0,04\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{ZnCl}_{2}\\
\\
\\
\end{array}$$\begin{array}{cc}
+ & \mbox{H}_{2}\\
& 0,04\mbox{ mol}\\
\\
\end{array}$$\begin{alignedat}{1}m_{\mbox{Zn}} & =n\times A_{r}\mbox{ Zn}\\
& =0,04\mbox{ mol}\times65\mbox{g/mol}\\
& =2,6\mbox{ g}\\
m_{\mbox{Cu}} & =5-m_{\mbox{Zn}}\\
& =5-2,6\mbox{ g}\\
& =2,4\mbox{ g}.
\end{alignedat}
$Salah$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{V_{\mbox{aktual}}}{V_{\mbox{teori}}}\times100\%\\
88,3\% & =\frac{800}{V_{\mbox{teori}}}\times100\%\\
V_{\mbox{teori}} & =960\mbox{ cm}^{3}\\
n_{\mbox{H}_{2}} & =\frac{960}{24000}\mbox{mol}\\
& =0,04\mbox{ mol}\\
\\
\end{alignedat}
$$\begin{array}{cc}
\mbox{Zn} & +2\mbox{HCl}\\
\frac{1}{1}\times0,04\mbox{ mol}\\
=0,04\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & \mbox{ZnCl}_{2}\\
\\
\\
\end{array}$$\begin{array}{cc}
+ & \mbox{H}_{2}\\
& 0,04\mbox{ mol}\\
\\
\end{array}$$\begin{alignedat}{1}m_{\mbox{Zn}} & =n\times A_{r}\mbox{ Zn}\\
& =0,04\mbox{ mol}\times65\mbox{g/mol}\\
& =2,6\mbox{ g}\\
m_{\mbox{Cu}} & =5-m_{\mbox{Zn}}\\
& =5-2,6\mbox{ g}\\
& =2,4\mbox{ g}.
\end{alignedat}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
1,3 gram gas etena dibakar sempurna dengan oksigen menghasilkan 3,3 gram gas $\mbox{CO}_{2}$. Tentukan volume gas oksigen yang dibutuhkan jika dalam pembakaran gas etena dihasilkan gas $\mbox{CO}_{2}$ 22,4 L (STP) ?
Betul$\begin{array}{c}
\mbox{C}_{2}\mbox{H}_{4}\\
1,3\mbox{ gram}\\
\frac{1,3}{26}\mbox{mol}\\
0,05\mbox{ mol}\\
\\
\end{array}$$\begin{array}{cc}
+ & 3\mbox{O}_{2}\\
\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CO}_{2}\\
\\
\,^{\,}\\
& \frac{2}{1}\times0,05\mbox{ mol}\\
& =0,1\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{alignedat}{1}V_{\mbox{aktual CO}_{2}} & =0,1\times44\mbox{ g}\\
& =4,4\mbox{ g}\\
\%\mbox{hasil} & =\frac{3,3\mbox{ gram}}{4,4\mbox{ gram}}\times100\%\\
& =75\%
\end{alignedat}
$Maka, untuk pembakaran kedua:
$\begin{aligned}n\mbox{ CO}_{2}\mbox{aktual} & =22,4\times\frac{1}{22,4}\mbox{ mol}\\
& =1\mbox{ mol}
\end{aligned}
$$\begin{aligned}75\% & =\frac{n\mbox{ CO}_{2}\mbox{aktual}}{n\mbox{ CO}_{2}\mbox{teoritis}}\times100\%\\
n\mbox{ CO}_{2}\mbox{teoritis} & =1,33\mbox{ mol}
\end{aligned}
$$\begin{array}{c}
\mbox{C}_{2}\mbox{H}_{4}\\
\\
\,^{\,^{\,}}\\
\\
\,^{\,^{\,}}
\end{array}$$\begin{array}{cc}
+ & 3\mbox{O}_{2}\\
\\
\\
& \frac{3}{2}\times1,33\mbox{ mol}\\
& =\mbox{2 mol}
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CO}_{2}\\
& 1,33\mbox{ mol}\\
\\
\\
\\
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\\
\end{array}$$\begin{alignedat}{1}V_{\mbox{O}_{2}} & =2\times22,4\mbox{L}\\
& =44,8\mbox{ L}.
\end{alignedat}
$Salah$\begin{array}{c}
\mbox{C}_{2}\mbox{H}_{4}\\
1,3\mbox{ gram}\\
\frac{1,3}{26}\mbox{mol}\\
0,05\mbox{ mol}\\
\\
\end{array}$$\begin{array}{cc}
+ & 3\mbox{O}_{2}\\
\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CO}_{2}\\
\\
\,^{\,}\\
& \frac{2}{1}\times0,05\mbox{ mol}\\
& =0,1\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}
\end{array}$$\begin{alignedat}{1}V_{\mbox{aktual CO}_{2}} & =0,1\times44\mbox{ g}\\
& =4,4\mbox{ g}\\
\%\mbox{hasil} & =\frac{3,3\mbox{ gram}}{4,4\mbox{ gram}}\times100\%\\
& =75\%
\end{alignedat}
$Maka, untuk pembakaran kedua:
$\begin{aligned}n\mbox{ CO}_{2}\mbox{aktual} & =22,4\times\frac{1}{22,4}\mbox{ mol}\\
& =1\mbox{ mol}
\end{aligned}
$$\begin{aligned}75\% & =\frac{n\mbox{ CO}_{2}\mbox{aktual}}{n\mbox{ CO}_{2}\mbox{teoritis}}\times100\%\\
n\mbox{ CO}_{2}\mbox{teoritis} & =1,33\mbox{ mol}
\end{aligned}
$$\begin{array}{c}
\mbox{C}_{2}\mbox{H}_{4}\\
\\
\,^{\,^{\,}}\\
\\
\,^{\,^{\,}}
\end{array}$$\begin{array}{cc}
+ & 3\mbox{O}_{2}\\
\\
\\
& \frac{3}{2}\times1,33\mbox{ mol}\\
& =\mbox{2 mol}
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CO}_{2}\\
& 1,33\mbox{ mol}\\
\\
\\
\\
\end{array}$$\begin{array}{cc}
+ & 2\mbox{H}_{2}\mbox{O}\\
\\
\,^{\,^{\,}}\\
\,^{\,^{\,}}\\
\\
\end{array}$$\begin{alignedat}{1}V_{\mbox{O}_{2}} & =2\times22,4\mbox{L}\\
& =44,8\mbox{ L}.
\end{alignedat}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Tembaga (II) nitrat mengalami dekomposisi berdasarkan reaksi berikut.
$2\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}(s)$ $\rightarrow2\mbox{CuO}(s)$ $+4\mbox{NO}_{2}(s)$ $+\mbox{O}_{2}(g)$
Berapakah massa $\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}$ jika dari reaksi diperoleh massa gas 20,4 gram dengan persentase hasil reaksi 94,4% ?
BetulMisal mol $\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}$$=2x\mbox{ mol}$
$\begin{array}{c}
\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}(s)\\
2x\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CuO}(s)\\
\\
\end{array}$$\begin{array}{cc}
+ & 4\mbox{NO}_{2}(g)\\
& 4x\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{O}_{2}(g)\\
& x\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{NO}_{2}} & =4x\mbox{ mol}\\
m_{\mbox{NO}_{2}} & =4x\times(14+16+16)\mbox{gram}\\
& =184x\mbox{ gram}\\
n_{\mbox{O}_{2}} & =x\mbox{ mol}\\
m_{\mbox{O}_{2}} & =x\times32\mbox{gram}\\
& =32x\mbox{ gram}\\
m_{\mbox{total gas}} & =184x+32x\\
& =216x\mbox{ gram}
\end{alignedat}
$$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\
94,4\% & =\frac{20,4}{216x}\times100\%\\
x & =0,1\\
n_{\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}} & =0,2\mbox{ mol}\\
m_{\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}} & =37,5\mbox{ gram}.
\end{alignedat}
$SalahMisal mol $\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}$$=2x\mbox{ mol}$
$\begin{array}{c}
\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}(s)\\
2x\mbox{ mol}
\end{array}$$\begin{array}{cc}
\rightarrow & 2\mbox{CuO}(s)\\
\\
\end{array}$$\begin{array}{cc}
+ & 4\mbox{NO}_{2}(g)\\
& 4x\mbox{ mol}
\end{array}$$\begin{array}{cc}
+ & \mbox{O}_{2}(g)\\
& x\mbox{ mol}
\end{array}$$\begin{alignedat}{1}n_{\mbox{NO}_{2}} & =4x\mbox{ mol}\\
m_{\mbox{NO}_{2}} & =4x\times(14+16+16)\mbox{gram}\\
& =184x\mbox{ gram}\\
n_{\mbox{O}_{2}} & =x\mbox{ mol}\\
m_{\mbox{O}_{2}} & =x\times32\mbox{gram}\\
& =32x\mbox{ gram}\\
m_{\mbox{total gas}} & =184x+32x\\
& =216x\mbox{ gram}
\end{alignedat}
$$\begin{alignedat}{1}\%\mbox{hasil} & =\frac{m_{\mbox{aktual}}}{m_{\mbox{teori}}}\times100\%\\
94,4\% & =\frac{20,4}{216x}\times100\%\\
x & =0,1\\
n_{\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}} & =0,2\mbox{ mol}\\
m_{\mbox{Cu}\left(\mbox{NO}_{3}\right)_{2}} & =37,5\mbox{ gram}.
\end{alignedat}
$