LiOH + CO2 → Li2CO3 + H2O?

Question

If the percent yield of the above reaction is found to be 90.02 %, how many grams of H2O are produced from the complete reaction of 71.85 g of LiOH ?

Answer 1

Mol. mass of lithium hydroxide = 23.95

71.85 g LiOH / 23.95 g/mol = 3.00 moles

Thus, with 100% yield, 3.00 moles of H2O would form, with a weight of 54 grams.

Taking into account the imperfect yield:

.9002 x 54 grams = 48.6108 grams H2O.

Answer 2

First you need to balance the equation. You need to add a 2 in front of the LiOH. That'll change the stoichiometry of the equation.

So, your theoretical yield would be:
(71.85 g/23.95 g/mol)X1mol H2O/2 mol LiOH X 18.01 g/mol = 27 g H2O.

If your actual yield is 90.02 %, just multiply the theoretical yield by 0.9002 to get 23.42 grams water