# 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?

Aug 31, 2014

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.

$P V = n R T$

$n = \frac{P V}{R T} = \left(101.0 \text{kPa"× 2.470"L")/(8.314 "kPa·L·K"^-1"mol"^-1 × 300"K}\right)$ = 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 $\frac{x}{65.4} + \frac{y}{24.3}$ = 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

$41.1 x = 80.442$

$x = \frac{80.442}{41.1} = 1.957$

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)