If 25.0 mL of methane gas react with 75.0 mL of oxygen gas,what is the volume of excess reactant (assume constant conditions)? CH₄(g)+ O₂(g)→ CO₂(g)+ H₂O(g)

Question

Quizplus · Accepted Answer

According to the balanced chemical equation, the stoichiometric ratio between methane and oxygen is 1:2. Therefore, for complete reaction, 50.0 mL of oxygen is required. However, only 75.0 mL of oxygen is present, which is in excess. The volume of excess reactant can be calculated by subtracting the volume of oxygen consumed in the reaction from the initial volume of oxygen. 
From the balanced chemical equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. So, the moles of oxygen required for the reaction = (25.0 mL of methane/1000 mL) x (1 L/1000 mL) x (1 mole CH4/1 L) x (2 moles O2/1 mole CH4) = 0.00005 moles of O2. 
Similarly, the moles of oxygen present = (75.0 mL of O2/1000 mL) x (1 L/1000 mL) x (1 mole O2/1 L) = 0.00015 moles of O2.
The moles of oxygen consumed in the reaction = 0.00005 moles of O2. 
Therefore, the volume of excess oxygen = [(initial volume of oxygen) - (volume of oxygen consumed)] = [(75.0 mL) - (25.0 mL)] = 50.0 mL.
Hence, the correct answer is C) 25.0 mL O₂.

If 250 ML of Methane Gas React with 75