Question

How many moles of excess reactant remains when $60.0mL$ of $1.0M$ hydrochloric acid reacts with $25.0mL$ of $0.80M$ barium hydroxide?

Answer 1

Number of moles of $HCl$

$n_1=1.0M\times \frac{60.0}{1000}L$

$n_1=0.06$ moles

$n_f$ of $HCl$ is $1$.

Number of moles of $Ba(OH{)}_2$

$n_2=0.80M\times \frac{25}{1000}L$

$n_2=0.02$ moles

$n_f$ of $Ba(OH{)}_2$ is $2$ .

Thus number of moles required of $HCl$ to react with $0.02$ moles of $Ba(OH{)}_2$ is

$\underset{\left(HCl\right)}{n_1\times 1}=0.02$ moles $\times 2$

$n_1=0.04$ moles of $HCl$ are required

Thus remaining moles of $HCl$ which are in excess is:

$\Rightarrow 0.06-0.04=0.02$ moles of $HCl$ .