A 3.66-g sample containing Zn (at.wt. 65.4) and Mg (24.3) reacted with a dilute acid to produce 2.470 L H2 gas at 101.0 kPa and 300 K. What is the percentage of Zn in the sample?
The sample contains 53.6 % Zn by mass.
Step 1. Write the balanced chemical equations.
Zn + 2HCl → ZnCl₂ + H₂
Mg + 2HCl → MgCl₂ + H₂
Step 2. Use the Ideal Gas Law to calculate the number of moles of H₂.
Step 3. Here's where it gets difficult. We set up and solve two simultaneous equations.
Let the mass of Zn =
Also, from the balanced equations, moles of Zn = moles of H₂ from Zn, and moles of Mg = moles of H₂ from Mg
Moles of H₂ = moles of Zn + moles of Mg =
Solve the two simultaneous equations for
Multiply by 65.4×24.3.
So we have 1.957 g of Zn
Step 3. Calculate the percentage of Zn.
% Zn =