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.

#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)