as given the equation is
2al + 6 hcl > 2 alcl3 + 3 h2(g)
as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas
given: three moles of al are present and 7 moles of hcl
for three moles of aluminum nine moles of hcl will be required
however as 7 moles of hcl are present so the limiting reagent is hcl
hcl is going to decide the number of moles of product formed
when six moles of hcl are present = 3 moles of h2 are formed
when one mole of hcl is present == 3/6 mole sof h2 will be formed
when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2
