Question

Zn + HCI > ZnCI2 + H2

A. What is my theoretical yield of H2 if I start with 74 g of HCI?

B. If my percent yield was 93%, what was my actual yield?

Answer 1

Solution.

1)

$Zn + 2HCl = ZnCl2 + H2$

$n(HCl) = \frac{m(HCl)}{M(HCl)}$

n(HCl) = 2.03 mol

$n(H2) = \frac{2.03}{2} = 1.015 \ mol$

$V(H2) = n(H2) \times Vm = 22.74 \ L$

2)

If we consider the percentage yield of zinc chloride, we find the theoretical mass of zinc chloride.

$n(ZnCl2) = \frac{2.03}{2} = 1.015 \ mol$

m(ZnCl2) = 138.04 g

$m(pract.) = y \times m(theory)$

m(pract.) = 128.38 g

Answer:

V(H2) = 22.74 L

m(pract.) = 128.38 g