Question

When 5.00 grams of Al are added to 305 mL of a 0.69 M solution of Hydrochloric acid, the products are a gas and an aqueous solution of an ionic compound. How many grams of Al are left unreacted?

Answer 1

2Al + 6HCl -----> 2AlCl3 + 3H2

no of moles of Al = 5/ 27 = 0.185 moles
no of moles of HCl = molarity * volume in L
                   = 0.69*0.305 = 0.21 moles
0.21 mole HCl*2moleAl/6 mole HCl = 0.07 moles Al.
So 0.21 mole HCl only needs 0.07 moles Al to react completely. We have much more
aluminium than that present so we can definitively say that HClis the limiting reagent.
if only 0.07 mole of Al react and we started with 0.185 moles of Al then left
over is 0.185-0.07 = 0.115 mle of Al
mass of Al = no of moles * gram atomic mass
           = 0.115*27 = 3.105 gm of Al