Question

Calculate the number of grams of aluminum sulfate that could sulfate be obtained by the action of 12.5 grams of aluminum on a excess of sulfuric acid.

2 Al + 3H2SO4 --> Al2(SO4)3 + 3H2

Answer 1

Molar masses are:

Al: 26.98 g/mol;

Al₂(SO₄)₃: 342.15 g/mol.

Therefore,

$m(Al_2(SO_4)_3)=12.5g(Al)\times\frac{1mol(Al)}{26.98g(Al)}\times\frac{1mol(Al_2(SO_4)_3)}{2mol(Al)}\times\frac{342.15g(Al_2(SO_4)_3)}{1mol(Al_2(SO_4)_3)}=79.3g$

Answer: 79.3 g