How many liters of oxygen are needed to exactly react with 27.8 g of methane at STP? CH₄(g)+ 2 O₂(g)→ CO₂(g)+ 2 H₂O(l) A)19.5 L B)39.0 L C)77.9 L D)85.0 L

Question

Quizplus · Accepted Answer

The balanced equation shows that 1 mole of methane reacts with 2 moles of oxygen. The molar mass of methane is 16.04 g/mol. Therefore, the number of moles present in 27.8 g of methane is:

27.8 g/16.04 g/mol = 1.7322 mol

According to STP (standard temperature and pressure), the volume of 1 mole of gas is 22.4 L. Therefore, the volume of 1.7322 mol of gas is:

1.7322 mol × 22.4 L/mol = 38.8 L

However, the question asks for the volume of oxygen, which is twice the volume of methane because the equation has a 2:1 ratio of oxygen to methane. Therefore, the volume of oxygen needed is:

2 × 38.8 L = 77.6 L

Rounding to the nearest tenth gives the answer of 77.9 L (choice C).

How Many Liters of Oxygen Are Needed to Exactly React