Question

How many liters of a .750 M solution of Hydrochloric acid are needed to react 27.6 grams of Zinc? (you will need to write a balanced equation). *

Answer 1

$Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2O$

Convert grams of Zn to moles of Zn:

$27.6 g Zn \times (\frac{1 mole Zn}{65.37 g Zn})=0.422 mol Zn$

Convert moles of Zn to moles of HCl:

$0.422 mol Zn \times (\frac{2 moles HCl}{1 mole Zn})=0.844 mol HCl$

As $c=\frac{n}{V}$ , then $V=\frac{n}{c} = \frac{0.844 mol HCl}{0.750 M}=1.13 L$