Hydrochloric Acid

Question

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. What volume of H2(g) is produced when 6.50 g of Al(s) reacts at STP?

Answer 1

Al + HCl → AlCl₃ + H₂

2Al + 6HCl → 2AlCl₃ + 3H₂

I wrote the equation and balanced it.  A balanced equation is required for deriving the mole ratio.

What volume of H2(g) is produced when 6.50 g of Al(s) reacts at STP?

6.50 g Al  x  (1 mole Al/26.98g Al)  x  3mole H₂/2mole Al = .361 mole H₂

 

.361 mole H₂  x  (22.4 L/1mole) = 8.09 L H₂