How many grams of CO2 are produced when 100 g C4H10 undergoes combustion with 200 g O2?
Use the following molar masses: C4H10=58g/mol, O2=32 g/mol and CO2=44g/mol.
Compare the moles of product formed from each reactant.
M(C4H10) = 58 g/mol;
M(O2) = 32 g/mol;
M(CO2) = 44/mol;
M(H2O) = 18 g/mol;
m(C4H10) = 100 g;
m(O2) = 200 g;
2C4H10 + 13O2 = 8CO2 + 10H2O;
n(C4H10) = m(C4H10)/M(C4H10) = 100/58 = 1.72 mol;
n(O2) = m(O2)/M(O2) = 200/32 = 6.25 mol.
Given the stoichiometric coefficients:
n(C4H10)' = n(C4H10)/2 = 0.86 mol;
n(O2)' = n(O2)/13 = 6.25/13 = 0.48 mol;
n(C4H10)' is higher than n(O2)', therefore C4H10 is excess and O2 is limited reagent.
For chemical reaction:
n(H2O) = 10/13 * n(O2) = 10 * 6.25/13 = 4.81 mol;
n(CO2) = 8/13 * n(O2) = 8 * 6.25/13 = 3.85 mol;
m(CO2) = n(CO2) * M(CO2) = 3.85 * 44 = 169.4 g.
Answer: 169.4 g.
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