CH4+2O2-->CO2+2H2O
If 22.75g of carbon dioxide is obtained when 9g of methane is burned in excess oxygen, what is the percent yield?
Show work too please
To find the percent yield, we need to compare the actual yield (the amount of carbon dioxide obtained) to the theoretical yield (the amount of carbon dioxide predicted by stoichiometry).
First, let's calculate the theoretical yield of carbon dioxide using stoichiometry:
1 mole of CH4 reacts to produce 1 mole of CO2.
So, we need to convert the mass of methane (9g) to moles:
Molar mass of CH4 = 12.01g (C) + 4(1.01g) (4H) = 16.05g/mol
Moles of CH4 = Mass of CH4 / Molar mass of CH4
= 9g / 16.05g/mol
≈ 0.56 mol
From the balanced equation, we know that 1 mole of CH4 produces 1 mole of CO2.
Therefore, the predicted moles of CO2 is also 0.56 mol.
Now, let's convert the predicted moles of CO2 to grams to determine the theoretical yield:
Molar mass of CO2 = 12.01g (C) + 2(16.00g) (2O) = 44.01g/mol
Mass of CO2 = Moles of CO2 * Molar mass of CO2
= 0.56 mol * 44.01g/mol
≈ 24.65g
So, the theoretical yield of carbon dioxide is approximately 24.65g.
Now, to find the percent yield:
Percent yield = (Actual yield / Theoretical yield) * 100
Given that the actual yield is 22.75g and the theoretical yield is 24.65g:
Percent yield = (22.75g / 24.65g) * 100
≈ 92.3%
Therefore, the percent yield of the reaction is approximately 92.3%.