In: Chemistry
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?
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