Question

The percent yield of a reaction between elemental zinc and an aqueous solution of 0.50 M hydro-chloric acid is known to be 78.0%. We need to produce 35.5 g of zinc chloride, what is the minimum amount in mL of hydrochloric acid that are required, given that zinc is in excess

Answer 1

Answer:

1.3 × 10³ mL

Explanation:

Let's consider the following reaction.

Zn + 2 HCl → ZnCl₂ + H₂

The percent yield is 78.0%. The real yield (R) of zinc chloride is 35.5 g. The theoretical yield (T) of zinc chloride is:

35.5 g (R) × (100 g T/ 78.0 g R) = 45.5 g T

The molar mass of zinc chloride is 136.29 g/mol. The moles corresponding to 45.5 g of zinc chloride is:

45.5 g × (1 mol/ 136.29 g) = 0.334 mol

The molar ratio of HCl to ZnCl₂ is 2:1. The moles of HCl that react with 0.334 moles of ZnCl₂ are 2 × 0.334 mol = 0.668 mol.

We need 0.668 moles of a 0.50 M HCl solution. The volume required is:

0.668 mol × (1000 mL/0.50 mol) = 1.3 × 10³ mL