Question

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

2 Al + 6HCl → 2 AlCl3 + 3 H2

Al is the limiting reactant, 9.0 mol H2 can be formed

HCl is the limiting reactant, 6.5 mol H2 can be formed

Al is the limiting reactant, 6.0 mol H2 can be formed

HCl is the limiting reactant, 4.3 mol H2 can be formed

Answer 1

compare the ratio that the balanced reaction tells you that you NEED with the amounts of reactants that the problem tells you that you HAVE\[\frac {6mol HCl}{2 mol Al} needed\]versus\[\frac{13mol HCl}{6mol Al} available\]

Answer 2

@JFraser is it HCl is the limiting reactant, 6.5 mol H2 can be formed

Answer 3

@JFraser or is it HCl is the limiting reactant, 4.3 mol H2 can be formed

Answer 4

in order to use all 6 moles of Al, you'll use 12 moles of HCl. that's what the balanced reaction tells you.

you have 13 moles of HCl, so the HCl won't run out first, the Al will.

Answer 5

@JFraser so then its Al is the limiting reactant, 6.0 mol H2 can be formed

Answer 6

knowing the Al runs out first lets you use the OTHER ratio, the ratio between the LR and the product you need to form:\[\frac{2mol Al}{3mol H_2} = \frac {6mol Al}{X mol H_2}\]