How many liters of oxygen are needed to exactly react with 25.8 g of methane at STP? CH₄(g)+ 2 O₂(g)→ CO₂(g)+ 2 H₂O(l) A) 18.1 L B) 36.2 L C) 72.3 L D) 78.9 L

Question

Quizplus · Accepted Answer

First, we need to balance the equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

According to the balanced equation, 1 mole of methane (16 g) reacts with 2 moles of oxygen (64 g). Therefore, 25.8 g of methane would require:

25.8 g CH4 × 1 mole CH4/16 g CH4 × 2 mole O2/1 mole CH4 × 22.4 L O2/ 1 mole O2 = 72.3 L

This is because at STP, one mole of ideal gas occupies 22.4 L.

How Many Liters of Oxygen Are Needed to Exactly React