Stoichiometry — degree of conversion from product analysis Carbon (6 g) is burned with air containing 18 g of O2. The products contain 16.5 g of CO2 and 2.8 g of CO (ignore other constituents). Based on disappearance of the limiting reactant, what is the degree of conversion?

Introduction / Context:
Product composition can be used to back-calculate conversion, especially when multiple products form (CO2 and CO). Here we determine which reactant is limiting, then compute how much of it reacted.

Given Data / Assumptions:

• m(C)in = 6 g; M(C) = 12 g/mol → n(C)in = 0.5 mol.
• m(O2)in = 18 g; M(O2) = 32 g/mol → n(O2)in = 0.5625 mol.
• Products: m(CO2) = 16.5 g; M(CO2) = 44 g/mol → n(CO2) = 0.375 mol.
• m(CO) = 2.8 g; M(CO) = 28 g/mol → n(CO) = 0.1 mol.

Concept / Approach:
Conversion of the limiting reactant = (moles reacted of limiting reactant)/(moles fed of limiting reactant). Each mole of CO2 or CO contains one mole of carbon; thus total moles of carbon that reacted equals n(CO2) + n(CO).

Step-by-Step Solution:
Compute carbon reacted: nC,reacted = 0.375 + 0.1 = 0.475 mol.Compute carbon fed: nC,in = 0.5 mol.Identify limiting reactant: compare O2 requirement for full CO2 (0.5 mol O2) with available 0.5625 mol → O2 in slight excess; therefore carbon is limiting.Degree of conversion = nC,reacted / nC,in = 0.475 / 0.5 = 0.95 → 95%.

Verification / Alternative check:
Oxygen consumed = for CO2: 0.375 mol O2; for CO: 0.5*0.1 = 0.05 mol O2; total = 0.425 mol O2 < 0.5625 mol available → consistent with O2 excess and carbon limiting.

Why Other Options Are Wrong:

• 100%: would require all 0.5 mol C to appear in products; only 0.475 mol reacted.
• 75% or 20%: inconsistent with calculated carbon in products.

Common Pitfalls:

• Forgetting that both CO2 and CO carry one carbon each when tallying reacted carbon.
• Misidentifying the limiting reactant by mass instead of moles.

Final Answer:
95%