Answer to Question #324636 in General Chemistry for Tesa

Question

What mass of ZnCl2 can be prepared from the reaction of 3.27 grams of zinc with 3.30 grams of HCl

Answer 1

Moles of zinc = $\dfrac{mass}{RAM} = \dfrac{3.27g}{65.38g/mol}$ = 0.05mol

The balanced equation for the reaction

Zn + 2HCl $\to$ ZnCl₂ + H₂

Mole ratios of Zn: HCl is 1:2

Moles of HCl required is therefore;

= $\dfrac{2}{1}x0.05 = 0.1mol$

But there are only = $\dfrac{3.30g}{36.458g/mol}=0.0905 moles of HCl$

Thus HCl is the limiting reagent

Moles of the product is therefore = $\dfrac{1}{2}x0.0905 = 0.045mol of ZnCl2$

Mass or product = moles x MM = 0.045mol x 136.286g/mol

= 6.168g