Question

a 6.11-g sample of Cu-Zn alloy reacts with HCl to produce hydrogen gas. if the hydrogen gas has a volume of 1.26 L, temperature of 22 degrees celcius and a pressure of 728 mm Hg. what is the percent of Zn in the alloy?

Answer 1

Two metals are in alloy, but only Zn reacts with HCl, Cu will not react with HCl, because it is less reactive than H₂.

Reaction between Zn and HCl

$$\mathrm{Zn} + 2\mathrm{HCl} = \mathrm{ZnCl_2} + \mathrm{H_2}$$

It is possible to find mass of Zn from this reaction and volume of H₂.

$\mathrm{PV} = \mathrm{nRT}$, where P is pressure (728 mmHg = 95.06 kPa), V volume (1.26L), R constant 8.31, T is temperature in Kelvin (22 + 273 = 295K), and n is amount.

$$n = \mathrm{PV} / \mathrm{RT} = 95.06 \times 1.26 / 8.31 \times 295 = 0.049 \mathrm{~mol}$$

n is also = m/Mw, where Mw is molecular mass of Zn, m is mass.

$$\mathrm{m} = \mathrm{Mw} \times \mathrm{n} = 65.38 \times 0.049 = 3.02 \mathrm{~g}$$

Now percent of Zn in alloy is:

$$\mathrm{w} = 3.02 / 6.11 \times 100\% = 49.47\%$$