6.11 g of a gold-zinc alloy reacts with hydrochloric acid.?

Molar mass of Au (gold) = 197 g/mol
Molar mass of Zn = 65.4 g/mol

For the H₂ gas produced :
Pressure, P = 728/760 atm
Volume, V = 0.00126 m³ = 1.26 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 22) K = 295 K

Gas law: PV = nRT
No. of moles of H₂ produced, n = PV/RT = (728/760) × 1.26 / (0.08206 × 295) = 0.0499 mol

Zinc reacts with HCl to give H₂, but gold does not.
Zn + 2HCl → ZnCl₂ + H₂
Mole ratio Zn : H₂ = 1 : 1
No. of moles of Zn in the alloy = No. of moles of H₂ produced = 0.0499 mol

Mass of Zn in the alloy = (0.0499 mol) × (65.4 g/mol) = 3.26 g
Mass of Au in the alloy = (6.11 - 3.26) g = 2.85 g
No. of moles of Au in the alloy = (2.85 g) / (197 g/mol) = 0.0145 mol

Mole fraction of Au in the alloy = 0.0145 / (0.0145 + 0.0499) = 0.225
Mole fraction of Zn in the alloy = 1 - 0.225 = 0.775

Since gold does not react with hydrochloric acid, zinc is metal that reacts with hydrochloric acid. Hydrogen is the gas that is produced. The following reaction occurs.

Zn + 2 HCl → ZnCl2 + H2

At standard temperature and pressure, the volume of one mole of a gas is 22.4 liters. One liter is 1000 cm^3

1 m = 100 cm
1 m^3 = 100^3 = 1 * 10^6 cm^3
V = 0.00126 * 1 * 10^6 = 1260 cm^3

This is 1.26 liters. Standard pressure is 760 mm Hg. Standard temperature is 273˚K
For this problem, T = 273 + 22 = 295˚K

P = V * n * R * T
P = 728 ÷ 760
(728 ÷ 760) * 1.26 = n * (22.4 ÷ 273) * 295
n * (22.4 * 295 * 760) = 728 * 1.26 * 273
n = 250,417.44 ÷ 5,022,080

The number of moles of hydrogen gas is approximately 0.05. This is also the number of moles of zinc.

Mass of one mole = 65.39 grams
Mass Zn = 65.39 * (250,417.44 ÷ 5,022,080)
This is approximately 3.26 grams. The rest is gold.

For gold, mass = 6.11 – [65.39 * (50,417.44 ÷ 5,022,080)]
This is approximately 2.85 grams.
Mass of one mole = 196.97 grams

n = [(6.11 – [65.39 * (50,417.44 ÷ 5,022,080)] ÷196.7
This is approximately 0.0014 mole.

Zn + 2HCl --> H2 + ZnCl2

use the ideal gas law. PV = nRT and solve for n. moles of H2 gas.
... P = 728, V = 0.00126m^3 or 1.26 L, R = 62.4, T = (273 + 22) = 295
... sub into equation,solve for n.

use stoich. calculations to find moles of Zn .. then use atomic mass to convert to grams of Zn.
... subtract .. 6.11 g alloy - g Zn (from above) = g Au
... convert mass Au to moles, use atomic mass
... now find mole fraction of Au .. moles Au / (moles Zn + moles Au)
... ... mole fr. Zn = moles Zn /(moles Zn + moles Au)

you should be able to go from here