Question

WILL GIVE MEDALS .
When 3.5 mol Al react with 12.5 mol HCl, what is the limiting reactant and how many moles of H can be formed?
2 Al + 6 HCl 2 AlCl3 + 3 H2

Answer 1

since 2mol of Al need 6mol of HCl, then\[3.5mol Al used * (6mol HCl \div 2mol Al) = 10.5mol HCl needed\] You have 12.5mol of HCl, but will only use 10.5mol. This makes the Al the LR. The yield is based off of using all the moles of the LR, so\[3.5mol Al (3mol HCl \div 2mol Al) = 5.25mol H{_2}\]