What is the mass percentage of aluminum in this alloy? I obtained the answers for Parts A, B, and C. Im confused on what to do in part D.?

Question

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H2(g) is collected over water at 29 ∘C and 752 torr, the volume is found to be 311 mL . The vapor pressure of water at 29 ∘C is 30.0 torr.

Part a) How many moles of H2 can be produced from x grams of Mg in magnesium-aluminum alloy? The molar mass of Mg is 24.31 g/mol.

Answer: .0411x mol H2

Part B) How many moles of H2 can be produced from y grams of Al in magnesium-aluminum alloy? The molar mass of Al is 26.98 g/mol.

Answer: .0556y mol H2

Part C)Taking into account the vapor pressure of water, how many moles of hydrogen gas, n, are present in 311 mL at 752 torr and 29 ∘C?The value of the gas constant R is 0.08206 L⋅atm/(mol⋅K). You may also find the conversion factors 1 atm=760 torr and TK=TC+273 useful.

Answer: n = 1.19×10−2 mol

Part D) What is the mass percentage of aluminum in this alloy?

Answer: ?

Answer 1

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Part a)

Mg + 2 HCl → MgCl2 + H2

(x g Mg) / (24.31 g Mg/mol) × (1 mol H2 / 1 mol Mg) = 0.0411353 x =

0.04114x  mol H2

Part B)

2 Al + 6 HCl → 2 AlCl3 + 3 H2

(y g Al) / (26.98 g Al/mol) × (3 mol H2 / 2 mol Al) = 0.0555967y = 0.05560y mol H2

Part C)

n = PV / RT = ((752 torr - 30.0 torr) / (760 torr/atm)) × (0.311 L) /

((0.08206 L⋅atm/(mol⋅K)) × (29 + 273) K) = 0.0119219 mol = 0.01192 mol H2

So you are correct about Parts a), B), and C).

Part D)

Given:

x + y = 0.250

From Parts a), B), and C):

0.0411353 x + 0.0555967 y = 0.0119219

Solve both equations simultaneously, using your favorite algebraic method:

x = 0.136728  and  y = 0.113272

(0.113272 g Al) / (0.250 g total) = 0.453088 = 45.3% Al by mass