Question

What is the mass, in grams, of aluminum metal required to produce 8.71 Liters of hydrogen gas at 35.7 ^{oC and 754 mm Hg according to the reaction:}

    2 Al   +   6 HCl ----->   2 AlCl₃   + 3 H₂

Answer 1

PV = nRT

P = 754 mmHg = 0.992 atm

V = 8.71 L

n = ?

R = Gas constant

T = 273 + 35.7 = 308.7 K

0.992 * 8.71 = n * 0.0821 * 308.7

8.64 = n * 25.3

n = 8.64 / 25.3 = 0.342 mole of H2

from the balanced equation we can say that

3 mole of H2 produced by 2 mole of Al so

0.342 mole of H2 will be produced by

= 0.342 mole of H2 *(2 mole of Al / 3 mole of H2 )

= 0.228 mole of Al

mass of 1 mole of Al = 27.0

so mass of 0.228 mole of Al = 6.16 g

Therefore, the mass of aluminum required would be 6.16 g