How many liters of oxygen are needed to exactly react with 19.8 g of methane at STP?
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

Question

How many liters of oxygen are needed to exactly react with 19.8 g of methane at STP? CH₄(g)+ 2 O₂(g)→ CO₂(g)+ 2 H₂O(l) A)13.9 L B)27.8 L C)55.5 L D)60.5 L

Quizplus · Accepted Answer

The balanced chemical equation shows that 1 mole of methane (CH4) reacts with 2 moles of oxygen gas (O2) to produce 1 mole of carbon monoxide (CO) and 2 moles of water (H2O).

To find out how many moles of oxygen gas are needed to react with 19.8 g of methane at STP, we can use the following calculations:

1. Calculate the number of moles of methane:

n = m/M
n = 19.8 g / 16.04 g/mol
n = 1.235 mol

2. Use the mole ratio from the balanced equation to calculate the number of moles of oxygen gas needed:

1 mol CH4 : 2 mol O2
1.235 mol CH4 : x mol O2

x = (1.235 mol CH4)(2 mol O2/1 mol CH4)
x = 2.47 mol O2

3. Calculate the volume of oxygen gas at STP using the molar volume of a gas (22.4 L/mol):

V = n x Vm
V = 2.47 mol x 22.4 L/mol
V = 55.4 L

Therefore, the answer is choice C) 55.5 L, which is the closest value to our calculated volume of oxygen gas needed to react with 19.8 g of methane at STP.

How Many Liters of Oxygen Are Needed to Exactly React