To ensure that the air in the shuttle remains free of excess CO2, engineers test the air purification system. They combine 1.000 × 103 g LiOH with 8.80 × 102 g CO2. The reaction produces 3.25 × 102 g H2O. If the reaction potentially could produce 360. g of water, what is the percent yield?
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To find the percent yield, we need to compare the actual amount of water produced to the theoretically maximum possible amount. The formula for percent yield is:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Given:
Mass of LiOH = 1.000 × 10^3 g
Mass of CO2 = 8.80 × 10^2 g
Mass of H2O (actual yield) = 3.25 × 10^2 g
Mass of H2O (theoretical yield) = 3.60 × 10^2 g
First, we need to determine the number of moles of LiOH, CO2, and H2O using their respective molar masses.
Molar mass of LiOH (Li = 6.941 g/mol, O = 16.00 g/mol, H = 1.008 g/mol):
LiOH = (6.941 g/mol) + (16.00 g/mol) + (1.008 g/mol) = 23.949 g/mol
Moles of LiOH = Mass of LiOH / Molar mass of LiOH = (1.000 × 10^3 g) / (23.949 g/mol)
Molar mass of CO2 (C = 12.01 g/mol, O = 16.00 g/mol):
CO2 = (12.01 g/mol) + (16.00 g/mol) + (16.00 g/mol) = 44.01 g/mol
Moles of CO2 = Mass of CO2 / Molar mass of CO2 = (8.80 × 10^2 g) / (44.01 g/mol)
Molar mass of H2O (H = 1.008 g/mol, O = 16.00 g/mol):
H2O = (1.008 g/mol) + (16.00 g/mol) = 18.02 g/mol
Moles of H2O (theoretical yield) = Mass of H2O (theoretical yield) / Molar mass of H2O = (3.60 × 10^2 g) / (18.02 g/mol)
Next, we need to find the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed. It is usually the reactant with the smallest number of moles compared to the stoichiometric ratio.
The balanced equation for the reaction is:
2 LiOH + CO2 → Li2CO3 + H2O
Looking at the stoichiometric ratio, we see that 2 moles of LiOH react with 1 mole of CO2 to produce 1 mole of H2O.
Using the mole ratios, we can calculate the number of moles of H2O that would be produced if all the LiOH reacted completely.
Moles of H2O (expected) = Moles of LiOH / 2
Now, we can calculate the theoretical yield of water (expected yield) in grams using the molar mass of H2O.
Mass of H2O (expected yield) = Moles of H2O (expected) * Molar mass of H2O
Finally, we can calculate the percent yield.
Percent Yield = (Actual Yield / Theoretical Yield) * 100 = (Mass of H2O (actual yield) / Mass of H2O (expected yield)) * 100