Question

If the theoretical yield of a reaction is 29.1 g and the actual yield is 20.5 g , what is the percent yield?

How many moles of H2 are formed by the complete reaction of 0.384 mol of Al?

Answer 1

Ans. # % yield = (Actual yield / Theoretical yield) x 100

                        = (20.5 g/ 29.1 g) x 100

                        = 70.45 %

# [[[Note: The question does not specify the what reacts with Al to produce H₂. It’s assumed that it is reaction of metal with dilute acid (say, HCl)- one of the commonest possible reaction. Please let me know if the reaction is a different one. ]]]

Balanced reaction:    2 Al + 6 HCl -----------> 2 AlCl₃ + 3 H₂

Stoichiometry: 2 mol of Al produces 3 mol H₂.

So, moles of H₂ produced by 0.384 mol Al = (3 mol H₂ / 2 mol Al) x 0.384 mol Al

                                                                        = 0.576 mol