Question

Carbon monoxide reacts with oxygen gas to form carbon dioxide according to the following chemical equation. 2CO(g)+O2(g)⟶2CO2(g) How many mL of oxygen gas will remain in the reaction vessel when 82.4mL of carbon monoxide gas reacts with 80.0mL of oxygen gas to form carbon dioxide under constant pressure and temperature?

Answer 1

Answer:

38.8 mL

Explanation:

2CO(g) + O₂(g)⟶2CO₂(g)

Because pressure and temperature remain constant, we can think of the moles ratios as volume ratios. This means that 2mL of CO react with 1 mL of O₂ to produce 2 mL of CO₂.

Because the problem asks us to calculate the amount of oxygen gas unreacted, then CO is the limiting reactant. We calculate the amount of O₂ that reacted, from the available amount of CO:

82.4 mL CO * $\frac{1mLO_{2}}{2mLCO}$ = 41.2 mL O₂

Thus, the oxygen gas remaining is:

80.0 mL - 41.2 mL = 38.8 mL O₂