10,8 gram alumunium padat (Ar = 27) direaksikan dengan larutan asam sulfat 0,2 M menghasilkan larutan alumunium sulfat dan gas hidrogen.
Reaksi
\(\ce{2 Al (s) + 3 H2SO4 (aq) -> Al2(SO4)3 (aq) + 3 H2 (g)}\)
Jumlah mol Al
\begin{equation*}
\begin{split}
n & = \frac {m}{Ar} \\\\
n & = \frac {10,8}{27} \\\\
n & = 0,4 \text{ mol}
\end{split}
\end{equation*}
Jumlah mol \(\ce{H2SO4}\)
\begin{equation*}
\begin{split}
n & = \frac {3}{2} \times 0,4 \\\\
n & = 0,6 \text{ mol}
\end{split}
\end{equation*}
Volume \(\ce{H2SO4}\)
\begin{equation*}
\begin{split}
n & = V \:.\: M \\\\
0,6 & = V \:.\: 0,2 \\\\
V & = 3 \text{ liter}
\end{split}
\end{equation*}
Jumlah mol \(\ce{H2}\)
\begin{equation*}
\begin{split}
n & = \frac {3}{2} \times 0,4 \\\\
n & = 0,6 \text{ mol}
\end{split}
\end{equation*}
Volume \(\ce{H2}\) pada kondisi RTP
\begin{equation*}
\begin{split}
n & = \frac {V}{24} \\\\
0,6 & = \frac {V}{24} \\\\
V & = 14,4 \text{ liter}
\end{split}
\end{equation*}
