Answer to Question #319405 in Physical Chemistry for tala

Question

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.

What is the limiting reactant?

what is the excess reactant?

how many grams of CO2 is produced?

Answer 1

m(C₄H₁₀) = 100 g;

M(C₄H₁₀) = 58 g/mol;

m(O₂) = 200 g;

M(O₂) = 32 g/mol;

M(CO₂) = 44 g/mol;

n(C₄H₁₀) = m(C₄H₁₀)/M(C₄H₁₀) = 100/58 = 1.72 mol;

n(O₂) = m(O₂)/M(O₂) = 200/32 = 6.25 mol;

2C₄H₁₀ + 13O₂ = 8CO₂ + 10H₂O;

By the chemical reaction:

n'(C₄H₁₀) = n(C₄H₁₀)/2 = 0.86 mol;

n'(O₂) = n(O₂)/13 = 0.48 mol;

So, O₂ is the limiting reactant and C₄H₁₀ is the excess reactant;

By the chemical reaction:

n(CO₂) = 8/13 * n(O₂) = 8/13 * 6.25 =3.85 mol;

m(CO₂) = n(CO₂) * M(CO₂) = 3.85 * 44 = 169.4 g.

Answer: O₂ is the limiting reactant;

C₄H₁₀ is the excess reactant;

169.4 g.