How many moles of hydrogen gas will be produced by the reaction of 4.05 g of aluminum with hydrochloric acid? Need help setting up the problem...is it:
4.05g AlCl3 x 1mol AlCl3/133.2 AlClc x 2 mol Al/3 mol H?
2 answers
first i usually write the problem over to see what i've got. remember the first rule is to turn moles into grams. finding out what moles you have and what grams you have. Divide moles into grams to solve for gram/moles. Cancel the moles so you're left with grams.dah! you've been shown and told this so many times isn't that what you are looking for if something else than you set it up as the end of what you are looking for till you get it. right!
It isn't 4.05g AlCl3 but 4.05g Al.
2Al + 6HCl ==> 2AlCl3 + 3H2
mols Al = 4.05g/atomic mass Al = ?
?mol Al x (3 mols H2/2 mols Al) = ?
2Al + 6HCl ==> 2AlCl3 + 3H2
mols Al = 4.05g/atomic mass Al = ?
?mol Al x (3 mols H2/2 mols Al) = ?