Question

. Which conversion factor should you use first to calculate the number of grams of CO2 produced by the reaction of 50.6 g of CH4 with O2? The complete combustion of methane is described by this equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) 44.0 g CO2 / 2 mol CO2 16.0 g CH4 / 1 mol CO4 1 mol CH4 / 16.0 g CH4 2 mol O2 / 1 mol CO2

Answer 1

Answer:

139.18 g.

Explanation:

• For the balanced equation:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l),

It is clear that 1 mol of CH₄ react with 2 mol of O₂ to produce 1 mol of CO₂ and 2 mol of H₂O.

• Firstly, we need to calculate the no. of moles of 50.6 g of CH₄:

no. of moles of CH₄ = mass/molar mass = (50.6 g)/(16.0 g/mol) = 3.1625 mol.

Using cross-multiplication:

1.0 mol of CH₄(g) produces → 1.0 mol of CO₂, from stichiometry.

∴ 3.1625 mol of CH₄(g) produces → 3.1625 mol of CO₂.

∴ The no. of grams of CO₂ produced = (no. of moles of CO₂)(molar mass of CO₂) = (3.1625 mol)(44.01 g/mol) = 139.18 g.