A 18.7 g sample of CaCO3 was treated with aqueous H2SO4 producing calcium sulfate, water and 3.65 g of CO2(g) according to:
CaCO3(s) + H2SO4(aq) -> CaSO4(s) + H2O + CO2(g).
What was the percent yield of CO2?
44.4
To calculate the percent yield of CO2, we need to compare the actual yield to the theoretical yield.
First, let's calculate the molar mass of CaCO3:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol) = 100.09 g/mol
Next, we need to calculate the number of moles of CaCO3:
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
moles of CaCO3 = 18.7 g / 100.09 g/mol = 0.186 mol
According to the balanced equation, 1 mole of CaCO3 reacts to produce 1 mole of CO2.
Therefore, the theoretical yield of CO2 is 0.186 mol.
To calculate the grams of CO2 in the theoretical yield:
mass of CO2 = moles of CO2 * molar mass of CO2
mass of CO2 = 0.186 mol * 44.01 g/mol = 8.175 g
Now, we can calculate the percent yield of CO2:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (3.65 g / 8.175 g) * 100 = 44.67%
Therefore, the percent yield of CO2 is approximately 44.67%.
To calculate the percent yield of CO2, we need to compare the actual yield of CO2 (3.65 g) to the theoretical yield of CO2.
First, we need to calculate the molar mass of CaCO3:
Ca: atomic mass = 40.08 g/mol
C: atomic mass = 12.01 g/mol
O: atomic mass = 16.00 g/mol
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + 3 × (16.00 g/mol) = 100.09 g/mol
Next, we need to determine the number of moles of CaCO3 that reacted:
Number of moles = mass / molar mass = 18.7 g / 100.09 g/mol = 0.1867 mol
Since the balanced equation shows that one mole of CaCO3 produces one mole of CO2, the theoretical yield of CO2 is also 0.1867 mol.
Now, we can calculate the theoretical mass of CO2:
Mass of CO2 = moles × molar mass = 0.1867 mol × 44.01 g/mol = 8.20 g
Finally, we can calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (3.65 g / 8.20 g) × 100 = 44.51%
Therefore, the percent yield of CO2 in this reaction is approximately 44.51%.
mols CaCO3 = grams/molar mass = ?
Using the coefficients in the balanced equation convert mols CaCO3 to mols CO2.
Now convert mols CO2 to grams CO2. g = mols x molar mass = ? This is the theoretical yield (TY). The actual yield (AY) in the problem is 3.65 grams.
%yield = (AY/TY)*100 = ?