When 10.7 g of Al are reacted with 42.5 g of HCl, what volume of H2 will be produced at 47 C and 725 mmHg?

Formula: 2Al+6HCl=2AlCl3+3H2

1) Write the balanced chemical equation:

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

2) Calculate the molar masses of Al and HCl and use them to convert the data to moles

Al: 27g/mol = > 10.7 gAl / 27g/mol = 0.396 mol Al

HCl: 1g/mol + 35.5 g/mol = 36.5g/mol = > 42.5gHCl / 36.5g/mol = 1.164 mol HCl

Theoretical ratio: 6 mol HCl / 2 mol Al = 3:1

Actual ratio: 1.164 mol HCl / 0.396 mol Al = 2.94 : 1

Then there is a little bit less HCl than the predicted by theoretical ratio, which means that this is the limitant reagent.

3) Use the amount of HCl to make the calculations of the proucts obtained.

Theoretical ratio: 3 mol H2 / 6 mol HCl = 1:3

1.164 mol HCl * 1 mol H2 / 3 mol HCl = 0.388 mol H2

4) Use the ideal gas formula to obtain the volume

pV = n RT

p = 725 mmHg * 1atm / 760mmHg = 0.954 atm

n = 0.388 mol

R = 0.082 atm*liter / K*mol

T = 47 °C + 273.15 = 320.15K

V = nRT/p = 0.388 mol * 0.082 [atm*liter/K*mol] * 320.15K / 0.954 atm = 10.7 liter

Answer: 10.7 liter