Question

A 20.00 g mixture of magnesium and zinc metal reacting with excess hydrochloric acid produced 1.212 g of hydrogen gas. What is the percent zinc in the original mixture?

Answer 1

No.of moles of H₂ produced = (1.212 g) / (2.016 g) = 0.60119 mol

We know that,

No.of moles of H₂ formed = No.of moles of Mg + No.of moles of Zn

Let X be the mass of Zn in mixture.

0.60119 mol = (20-X) / (24.31 g/mol) + (X/65.38 g/mol)

Solving for X, we get:

X = 8.572564 g

Therefore, Mass of Zn in mixture = 8.572564 g

Hence, Mass percentage of Zn in the mixture is given by:

Mass % = (Mass of Zn) / (total mass) * 100%

Mass % = (8.572564 g) /(20 g) * 100%

Mass % = 0.4286 * 100%

Mass % = 42.86 % ------------------- (**Answer)