The reactions of carboxylic acids
Exercise 2
$50\;mL$ of a $0.20\;M$ solution of formic acid react with $0.49\;g$ zinc.
Calculate the molarity of ions $Zn^{2+}$ at the end of the reaction.
$n_{acid}$ $=$ $0.20\cdot 0.050$ $=$ $0.010\; mol$
$n_{Zn}$ $=$ $\frac{0.49}{65.4}$ $=$ $0.0075\; mol$
Simplified equation:
$Zn$ $+$ $2CH_3COOH$ $\rightarrow$ $Zn^{2+}$ $+$ $2CH_3COO^-$ $+$ $H_2$
Stoechiometric proportions:
$1\; mol$ $+$ $2\; mol$ $\rightarrow$ $1 \;mol$ $+$ $2\;mol$ $+$ $1\; mol$
Limiting agent: Formic acid (see the stoechiometric proportions!):
$0.010\;mol$ acid give $0.005\;mol$ $Zn^{2+}$
$[Zn^{2+}]$ $=$ $\frac{n}{V}$ $=$ $\frac{0.005}{0.05}$ $=$ $0.10 \frac{mol}{L}$
