# 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?

##### 1 Answer

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 =

Then **Equation 1**)

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 =

So **Equation 2**)

Solve the two simultaneous equations for

From Equation1,

So

Multiply by 65.4×24.3.

So we have 1.957 g of Zn

**Step 3.** Calculate the percentage of Zn.

% Zn =