Answer to Question #105983 in General Chemistry for george

Question

Calculate the mass of hydrogen formed when 3.79 g of aluminum reacts with excess hydrochloric acid according to the balanced equation: 2Al + 6 HCl → 2 AlCl3 + 3 H2
molar masses
Al = 26.98 g/mol HCl = 36.46 g/mol H2 = 2.02 g/mol AlCl3 = 133.33 g/mol

Answer 1

2Al + 6HCl → 2AlCl₃ + 3 H₂ 

The mole ratio between aluminium and the hydrogen produced is 2:3

$moles =\frac{mass}{RFM}$

3.79g of Al= 3.79/26.98 =0.1405 moles

If 0.1405 moles of Al reacted then 0.2108 moles of hydrogen were produced.

0.2108 moles of hydrogen =2.02*0.2108=0.43g