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?
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) = ?
To determine the number of moles of hydrogen gas produced by the reaction of 4.05 g of aluminum with hydrochloric acid, we need to set up a balanced chemical equation and use stoichiometry.
The balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl) is:
2Al + 6HCl -> 2AlCl3 + 3H2
From the balanced equation, we can see that for every 2 moles of aluminum, 3 moles of hydrogen gas are produced.
Now, let's set up the problem:
1. Convert the mass of aluminum (4.05 g) to moles.
To do this, we use the molar mass of aluminum (Al), which is 26.98 g/mol.
moles of Al = mass of Al / molar mass of Al
= 4.05 g / 26.98 g/mol
2. Determine the number of moles of hydrogen gas produced using stoichiometry.
Based on the balanced equation, we know that 2 moles of Al produce 3 moles of H2.
moles of H2 = moles of Al * (3 moles H2 / 2 moles Al)
So, the correct setup for the problem is:
moles of H2 = (4.05 g Al / 26.98 g/mol) * (3 mol H2 / 2 mol Al)
= 0.2997 mol H2
Therefore, approximately 0.2997 moles of hydrogen gas will be produced by the reaction.