The equation for the complete combustion of methane is:
CH4(g) + 2O2(g) CO2 (g) +2H2O (g)
If 1.60 g of methane were completely burned in 8.00 g of oxygen (an excess) to produce 4.40 g
of carbon dioxide, what mass of oxygen is left unreacted?
(Ar: H = 1 ; C = 12 ; O = 16)
Answer:
26) B
The main formula that needs to be used is: moles = (mass ÷ Mr)
n = m ÷ Mr
Mr of CH₄ = 12 + (1 x 4) = 16
Mr of CO₂ = 12 + (16 x 2) = 44
Mr of O₂ = 16 x 2 = 32
Moles of CH₄ = (1.6 ÷ 16) = 0.1 mol
Moles of CO₂ = (4.4 ÷ 44) = 0.1 mol
Moles of O₂ = (8 ÷ 32) = 0.25 mol
However, the molar ratio O₂ and CO₂ = 2 : 1
Mole of CO₂ is = 0.1 mol
Therefore, Moles of O₂ used = 0.1 x 2 = 0.2 mol
Moles of unreacted O₂ = 0.25 – 0.20 = 0.05 mol
Moles = Mass ÷ Mr
Mass = Moles x Mr
Mass = 0.05 x 32 = 1.6g
