# Zinc oxide reacts with hydrochloric acid to produce zinc chloride and dihydrogen monoxide. If 3.6 moles of zinc oxide reacts with an excess of hydrochloric acid, then how many grams of zinc chloride were produced?

Jan 10, 2016

$490.7 \text{g } Z n C {l}_{2}$

#### Explanation:

The reaction happening in solution is the following:

$Z n O \left(a q\right) + 2 H C l \left(a q\right) \to Z n C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right)$

Since the hydrochloric acid ($H C l$) is in excess, therefore, the number of moles of Zinc oxide ($Z n O$) will determine the mass of Zinc chloride ($Z n C l$).

Using dimensional analysis, we find:

?g ZnCl_2=3.6cancel("mol " ZnO)xx(1 cancel ("mol "ZnCl_2))/(1cancel("mol " ZnO))xx(136.3"g "ZnCl_2)/(1 cancel ("mol "ZnCl_2))=490.7"g "ZnCl_2