Considering the limiting reactant,what is the volume of the excess reactant that remains after the reaction of 25.0 mL of methane gas and 75.0 mL of oxygen gas? (Assume constant conditions. ) CH₄(g)+ 2 O₂(g)→ CO₂(g)+ 2 H₂O(g)

Question

Quizplus · Accepted Answer

To determine the limiting reactant, we need to calculate the amount of product each reactant can produce. 
First, we need to convert the given volumes of reactants to moles using the ideal gas law:

n CH₄ = V/VM = 25.0 mL/ 22.4 L/mol = 0.00112 mol 
n O₂ = V/VM = 75.0 mL/ 22.4 L/mol = 0.00335 mol

Next, we can use the balanced chemical equation to see how much product each reactant can produce:

1 mol CH₄ produces 1 mol CO₂ 
0.00112 mol CH₄ produces 0.00112 mol CO₂

1 mol O₂ produces 2 mol CO₂
0.00335 mol O₂ produces 0.00670 mol CO₂

Since methane can only produce 0.00112 mol of CO₂ while oxygen can produce 0.00670 mol of CO₂, oxygen is the excess reactant.

To determine the volume of the excess reactant, we can use the ideal gas law again:

V = n x VM = 0.00335 mol x 22.4 L/mol = 0.075 L or 75 mL

Therefore, 25.0 mL of methane reacts completely and 25.0 mL of oxygen remains as the excess reactant. 
The correct answer is choice C) 25.0 mL O₂.

Considering the Limiting Reactant,what Is the Volume of the Excess