The following equation represents the partial combustion of methane,CH₄.
2CH₄(g) + 3O₂(g) $\to$ 2CO(g) + 4H₂O(g)
At constant temperature and pressure,what is the maximum volume of carbon monoxide that can be obtained from 3.39 $\times$ 10² L of methane and 1.70 $\times$ 10² L of oxygen?

Question

The following equation represents the partial combustion of methane,CH₄.
2CH₄(g) + 3O₂(g)

\to

2CO(g) + 4H₂O(g)
At constant temperature and pressure,what is the maximum volume of carbon monoxide that can be obtained from 3.39

\times

10² L of methane and 1.70

\times

10² L of oxygen?

Quizplus · Accepted Answer

According to the given equation, 1 mole of methane produces 2 moles of CO. 
Number of moles of methane = (3.39 x 10^10)/22.4 = 1.515 x 10^9 
Number of moles of oxygen = (1.70 x 10^10)/22.4 = 7.589 x 10^8 (oxygen is limiting reagent) 
Using the mole ratio, number of moles of CO produced = 2 x 7.589 x 10^8 = 1.518 x 10^9 
Volume of CO produced = (1.518 x 10^9 x 22.4)/1000 = 33.96 x 10^6 L = 1.13 x 10^11 L 
The answer is therefore D, 1.13 x 10^10 L.

The Following Equation Represents the Partial Combustion of Methane,CH4 →\to→ 2CO(g)+ 4H2O(g) at Constant Temperature and Pressure,what Is the Maximum

The Following Equation Represents the Partial Combustion of Methane,CH₄ $\to$ 2CO(g)+ 4H₂O(g)
at Constant Temperature and Pressure,what Is the Maximum