Question

How many liters of hydrogen gas would be produced by the reaction of 40.4 g of Al with excess HCl at STP according to the following reaction? 2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

Answer 1

The balanced equation is

2 Al + 6Hcl --------------> 2 AlCL3 + 3H2

40.4g/27g/mol=1.496 xs - - initial moles

0 xs 1.496 1.496x3/2 after rxn

moles of H2 formed = 1.496x3/2 = 2.244 mol

At STP 1 mole of any gas occupies 22.4L

2.244 moles of H2 occupies volume = 2.244molx22.4L/ 1 mol =50.275 L