Sebanyak 2 gram serbuk kalsium karbonat direaksikan dengan 100 mL larutan asam klorida 2 M menghasilkan kalsium klorida, gas karbon dioksida dan air.
Reaksi
\(\ce{CaCO3 (s) + 2 HCl (aq) -> CaCl2 (aq) + CO2 (g) + H2O (l)}\)
Menentukan mol zat mula-mula
\(\ce{mol CaCO3}\)
\begin{equation*}
\begin{split}
n & = \frac {m}{Mr} \\\\
n & = \frac {2}{40 + 12 + 3 \:.\: 16} \\\\
n & = 0,02 \text{ mol}
\end{split}
\end{equation*}
\(\ce{mol HCl}\)
\begin{equation*}
\begin{split}
n & = V \:.\: M \\\\
n & = 0,1 \:.\: 2 \\\\
n & = 0,2 \text{ mol}
\end{split}
\end{equation*}
Tabel reaksi
\begin{equation*}
\begin{array}
& & \ce{CaCO3 (s) & + & 2 HCl (aq) & -> & CaCl2 (aq) & + & CO2 (g) & + & H2O (l)} \\
\text{Mula-mula} & 0,02 && 0,2 && - && - && - \\
\text{Reaksi} & 0,02 && 0,04 && 0,02 && 0,02 && 0,02 \\
\text{Sisa} & - && 0,16 && 0,02 && 0,02 && 0,02
\end{array}
\end{equation*}
Pereaksi pembatas
\(\ce{CaCO3}\)
Sisa reaktan
\(\ce{0,16 mol HCl}\)
\(\ce{Volume CO2}\)
\begin{equation*}
\begin{split}
n & = \frac {V}{22,4} \\\\
0,02 & = \frac {V}{22,4} \\\\
V & = 0,448 \text{ liter}
\end{split}
\end{equation*}
