Question

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?

Answer 1

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.

`PV = nRT`

`n = (PV)/(RT) = (101.0"kPa"× 2.470"L")/(8.314 "kPa·L·K"^-1"mol"^-1 × 300"K")` = 0.1000 mol

Step 3. Here's where it gets difficult. We set up and solve two simultaneous equations.

Let the mass of Zn = `x` g and the mass of Mg = `y` g.

Then `x + y` = 3.66 (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 = `x" g Zn" × (1"mol Zn")/(65.4"g Zn") + y" g Mg" × (1"mol Mg")/(24.3"g Zn")`

So `x/65.4 + y/24.3` = 0.1000 (Equation 2)

Solve the two simultaneous equations for `x` and `y`.

From Equation1, `y = 3.66 - x`

So `x/65.4 + (3.66 – x)/24.3 = 0.1000`

Multiply by 65.4×24.3.

`24.3x + 65.4(3.66 – x)= 0.1000 × 65.4 × 24.3`

`24.3x + 239.364 – 65.4 x = 158.922`

`41.1x = 80.442`

`x= 80.442/41.1 = 1.957`

So we have 1.957 g of Zn

Step 3. Calculate the percentage of Zn.

% Zn = `"mass of Zn"/"mass of sample" × 100 % = (1.957"g")/(3.66"g") × 100 % = 53.5 %` (m/m)