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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.?
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?
- Roger the MoleLv 72 months agoFavorite Answer
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
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
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).
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