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?
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)
A 20.00 g mixture of magnesium and zinc metal reacting with excess hydrochloric acid produced 1.212...
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