Question

Consider the balanced chemical equation for the combustion of methane (CH4).

mc026-1.jpg

Given that the molar mass of CO2 is 44.01 g/mol, how many liters of oxygen is required at STP to produce 88.0 g of CO2 from this reaction?
44.8 L
45.00 L
89.55 L
89.6 L

Answer 1

Took a guess and got it right!

D) 89.6 L

Answer 2

89.6 L of O₂

Solution:

The balanced chemical equation is as,

CH₄ + 2 O₂ → CO₂ + 2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂

So,

88 g CO₂ will be produced by = X L of O₂

Solving for X,

X = (88 g × 44.8 L) ÷ 44 g

X = 89.6 L of O₂