How much hydrogen gas (H2) forms when 54 g of aluminum reacts with excess HCI ?

Answer 1

first we write the chemical reaction:
Al + HCl -> AlCl3 + H2
then we balance it:
2Al + 6HCl -> 2AlCl3 + 3H2
also we need to look for the masses of Al. from the periodic table,
Al = 27 g/mol
therefore,
54 g Al * (1 mol Al / 27 g Al) * (3 mol H2 / 2 mol Al) = 3 mol H2
or in grams, (atomic mass of H = 1 g/mol)
3 mol H2 * (2 g H2 / 1 mol H2) = 6 g H2

hope this helps~ :)