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 * [tex]\frac{1mLO_{2}}{2mLCO}[/tex] = 41.2 mL O₂
Thus, the oxygen gas remaining is:
80.0 mL - 41.2 mL = 38.8 mL O₂
The volume of oxygen gas, O₂, remaining in the reaction vessel when 82.4 mL of carbon monoxide gas reacts with 80.0 mL of oxygen gas is 38.8 mL
From the question given above, we can conclude that CO is the limiting reactant since the question is asking for the volume of O₂ remaining.
Now, we shall obtain the volume of O₂ that reacted. This can be obtained as follow:
2CO(g) + O₂(g) —> 2CO₂(g)
From the balanced equation above,
2 mL of CO reacted with 1 mL of O₂.
Therefore,
82.4 mL of CO will react with = [tex]\frac{82.4}{2}\\\\[/tex] = 41.4 mL of O₂.
Finally, we shall determine the volume of O₂ remaining. This can be obtained as follow:
Volume of O₂ given = 80 ml
Volume of O₂ that reacted = 41.4 mL
Volume of O₂ remaining = (Volume of O₂ given) – (Volume of O₂ that reacted)
Volume of O₂ remaining = 80 – 41.2
Thus, the volume of O₂ remaining in the reaction vessel is 38.8 mL
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