Question

When 4.5 mol Al react with 11.5 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

2Al + 6HCl → 2AlCl3 + 3H2



Al is the limiting reactant, and 4.5 mol H2 can be formed

HCl is the limiting reactant, and 3.8 mol H2 can be formed

Al is the limiting reactant, and 6.8 mol H2 can be formed

HCl is the limiting reactant, and 5.8 mol H2 can be formed

Answer 1

5.8 mol is the answer

Answer 2

1)By looking at the balanced equation, you can see that 2 mol of Al needs 6 moles of HCl to fully react . So to react 4.5 moles of Al, you need x of HCl.

2 mol of Al ----------6moles of HCl
4.5 mol of Al -----------x
moles of HCl needed = 13,5 moles HCl
However, you don't have 13,5 moles, you only have 11.5moles - This means the HCl is the limiting reagent.

2)
Using the same method, 6 mol of HCl originates 3 moles of H2
So, if we only have 11,5 mol of the limiting reactant to react, how much of H2 will form as a product?
6 mol of HCl----------3 moles of H2
11,5 mol of HCl -----------x
x= (11,5 x 3)/ 6
x= 5,75 mol of H2 (approximately 5,8 mol)
Answer: D