Question

Zinc and hydrochloric acid react according to the equation
$Zn\left(s\right)+2HCl\left(aq\right)\to ZnC{l}_2\left(aq\right)+H_2\left(g\right)$
0.30 mole of $Zn$ is added to 0.52 mole of $HCl$. Which of the two reactants is the limiting reagent and how many moles of $H_2$ is produced?

• A. $Zn,0.26\text{mol}$
• B. $HCl,0.30\text{mol}$
• C. $HCl,0.26\text{mol}$
• D. $Zn,0.30\text{mol}$

Answer 1

The correct option is C $HCl,0.26\text{mol}$

The given reaction is:
$Zn+2HCl\to ZnC{l}_2+H_2$
1 mole of $Zn$ reacts with 2 mol of $HCl$ to produce 1 mol of $H_2$
0.3 mol of $Zn$ will react with 0.6 mol of $HCl$ but there is only 0.52 mol of $HCl$.
So, $HCl$ is the limiting reagent.
2 mol of $HCl$ gives 1 mol of $H_2$
So, 0.52 mol of $HCl$ will give 0.26 mol of $H_2$