50 ml of 5 M H2SO4 react with excess BaCl2 to Make 40 g of BaSO4.
a.) What is the theoretical yield of BaSO4?
b.) What is the Percentage yield of BaSO4
Here is an example stoichiometry problem I've posted. Just follow the steps to find the theoretical yield and the percent yield.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the theoretical yield of BaSO4, we need to determine the limiting reagent in the chemical reaction and calculate the stoichiometric ratio between the limiting reagent and the product. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
In this case, we know that 50 ml of 5 M H2SO4 reacts with excess BaCl2, so the H2SO4 is the limiting reagent. To calculate the moles of H2SO4, we use the formula:
moles = concentration * volume (in liters)
moles of H2SO4 = 5 M * 0.05 L = 0.25 moles
From the balanced chemical equation, we know that the stoichiometric ratio between H2SO4 and BaSO4 is 1:1. Therefore, the moles of BaSO4 formed will also be 0.25 moles.
Now, to calculate the theoretical yield of BaSO4 in grams, we need to use its molar mass which is 233.39 g/mol:
theoretical yield = moles of BaSO4 * molar mass of BaSO4
= 0.25 moles * 233.39 g/mol
= 58.35 g
So, the theoretical yield of BaSO4 is 58.35 grams.
To calculate the percentage yield of BaSO4, we need to compare the obtained yield (actual yield) of 40 grams with the theoretical yield we just calculated.
percentage yield = (actual yield / theoretical yield) * 100
= (40 g / 58.35 g) * 100
= 68.58%
Therefore, the percentage yield of BaSO4 is approximately 68.58%.