Asked by Veronique
The balanced reaction equation for the combustion of heptane, C7H16, is given below. If the reaction produced 37.5 g CO2, how many grams of heptane were burned?
Answers
Answered by
DrBob222
C7H16 + 11O2 ==> 7CO2 + 8H2O
mols CO2 = grams/molar mass = 37.5/44 = approx 0.85
Convert mols CO2 to mols C7H16 using the coefficients in the balanced equation.
approx 0.85 x (1 mol C6H16/8 mols CO2) = approx 0.85 x 1/8 = 0.1 mol C7H16
g C7H16 = mols C7H16 x molar mass C7H16 = ? grams.
Check my work. It's late.
mols CO2 = grams/molar mass = 37.5/44 = approx 0.85
Convert mols CO2 to mols C7H16 using the coefficients in the balanced equation.
approx 0.85 x (1 mol C6H16/8 mols CO2) = approx 0.85 x 1/8 = 0.1 mol C7H16
g C7H16 = mols C7H16 x molar mass C7H16 = ? grams.
Check my work. It's late.
Answered by
Bob frick you
Frick you bob you lying basterd!
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