Answer:
443 L of carbon dioxide
Step-by-step explanation:
Combustion reaction: CH₄ (g) + 2O₂ (g) → CO₂(g) + 2H₂O(g)
We assume, that the oxygen is the excess reagent, so let's convert the mass of methane to moles. (mass (g) / molar mass)
Firstly we need to convert the mass from kg to g → 0.300 kg .1000 g / 1kg
300 g / 16 g/mol = 18.75 moles of methane.
Ratio is 1:1, so 18.75 moles of methane will produce 18.75 moles of CO₂
We apply the Ideal Gases Law to find the answer:
Firstly we need to convert the T°C to T°K → 15°C + 273 = 288 K
P . V = n . R. T → V = ( n . R . T ) / P We replace data:
V = (18.75 mol . 0.0821 L.atm/mol.K . 288K) / 1 atm → 442.8 ≅ 443 L