The reaction of 9.58 g of Carbon with excess O2 yields 8.90 g of CO2, what is the percent yield of this reaction?
C + O2 ==> CO2
mols C = g/atomic mass C = estimated 0.3
mols CO2 = 0.3 since there is 1 mol CO2 for each mol C.
g CO2 = mols CO2 x molar mass CO2 = ? and that is the theoretical yield (TY). The problem tells you the actual yield is 8.90g.
%yield = [AY/TY]*100 = ?
98
To calculate the percent yield of a reaction, we need to compare the actual yield (what was obtained in the experiment) with the theoretical yield (what should have been obtained according to stoichiometry). The formula to calculate percent yield is:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
In this case, we have the mass of the reactant (carbon) and the mass of the product (CO2). We can use these values to determine the actual yield of CO2.
First, let's calculate the theoretical yield of CO2. To do this, we need to find the balanced chemical equation for the reaction between carbon and oxygen to form carbon dioxide.
Assuming that the reaction is:
C + O2 → CO2
By looking at the balanced equation, we can see that for each mole of carbon (12.01 g), we should obtain one mole of CO2 (44.01 g). Therefore, we can use the molar masses of carbon and CO2 to convert the mass of carbon into the theoretical yield of CO2.
The molar mass of carbon is 12.01 g/mol, and the molar mass of CO2 is 44.01 g/mol.
(Convert mass of carbon to moles) Moles of carbon = Mass of carbon / Molar mass of carbon
Moles of carbon = 9.58 g / 12.01 g/mol = 0.798 moles
Since the stoichiometry of the reaction is 1:1 (one mole of carbon reacts with one mole of CO2), the theoretical yield of CO2 is also 0.798 moles.
Now, we can calculate the theoretical mass of CO2:
Theoretical Mass of CO2 = Moles of CO2 x Molar mass of CO2
Theoretical Mass of CO2 = 0.798 moles x 44.01 g/mol = 35.05 g
Finally, we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Percent Yield = (8.90 g / 35.05 g) x 100 ≈ 25.37%
Therefore, the percent yield of this reaction is approximately 25.37%.