Question

If 0.032g impure sample of aluminum is reacted with an excess of hydrochloric acid producing 32.4 mL of hydrogen gas, what is the percent purity of the aluminum if the hydrogen gas was collected over water at a temperature of 24.0 celcius and a pressure of 742 mmHg

Answer 1

1st calculate the mol of H2

Given:

P = 742.0 mm Hg

= (742.0/760) atm

= 0.976 atm

V = 32.4 mL

= (32.4/1000) L

= 0.0324 L

T = 24.0 oC

= (24.0+273) K

= 297 K

find number of moles using:

P * V = n*R*T

0.976 atm * 0.0324 L = n * 0.0821 atm.L/mol.K * 297 K

n = 1.297*10^-3 mol

This is number of mol of H2 produced

Balanced chemical reaction taking place is:

2Al + 6 HCl —> 2 AlCl3   +   3H2

so,

mol of Al required = (2/3)*mol of H2

= (2/3)*1.297*10^-3 mol

= 8.647*10^-4 mol

Lets calculate the mass of Al reacted

molar mass of Al = 26.98 g/mol

mass of Al reacted = mol of Al * molar mass

= (8.647*10^-4 mol)*(26.98 g/mol)

= 0.023 g

This is mass of pure Al

% purity = 0.023*100/0.032

= 72 %

Answer: 72 %