If 16.9 g of CO2 was produced from the thermal decomposition of 41.83 g of CaCO3, what is the percentage yield of the reaction?
CaCO3(s) > CaO(s) + CO2(g)
Calculate the theoretical yield
41.83 g of CaCO3 *(1 mole/100.08 g of CaCO3)= moles of CaCO3
SInce the mole ratios are equal, moles of CaCO3=CO2 moles of CO2
moles of CO2*(44.01 g of CO2/moles of CO2)= g of CO2 from the theoretical reaction.
(16.9 g of CO2 was produced/g of CO2 from theoretical rxn)*100= % yeild
Thanks!
To calculate the percentage yield of a reaction, we need to compare the actual yield with the theoretical yield.
First, let's determine the theoretical yield, which is the maximum amount of CO2 that could be produced from the given amount of CaCO3.
1 mole of CaCO3 produces 1 mole of CO2. We'll find the number of moles of CaCO3 using its molar mass (40.08 g/mol for CaCO3):
41.83 g CaCO3 × (1 mol CaCO3 / 100.09 g CaCO3) = 0.4178 mol CaCO3
Since the molar ratio between CaCO3 and CO2 is 1:1, the number of moles of CO2 produced will be the same.
Now, we need to calculate the actual yield, which is given as 16.9 g CO2.
By using the molar mass of CO2 (44.01 g/mol), we can convert the mass of CO2 to moles:
16.9 g CO2 × (1 mol CO2 / 44.01 g CO2) = 0.384 mol CO2
Now we can calculate the percentage yield:
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (0.384 mol CO2 / 0.4178 mol CO2) × 100 = 92.05%
Therefore, the percentage yield of the reaction is approximately 92.05%.