$2CO(g) + O_2(g) -> 2CO_2(g)$. How many liters of $O_2$ will react with 74.2 liters of CO? If 6.5 g of CO react, how many liters of $CO_2$ are produced?

Md Ariful Islam

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Md Ariful Islam

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We calculate the relative quantities from the molar ratios in the balanced reaction equation.

$2CO(g)+O_2(g)→2CO_2(g)$.
As a pretty good approximation, gas volumes are proportional to molar amounts. Therefore with a molar ratio of 2:1 of $CO : O_2$ we can calculate that 74.2 L of CO will react with
$(74.2/2) = 37.1$ L of $O_2$.

Similarly, converting the CO mass to moles, we have $(6.5g)/(28"g/mol") = 0.232$ moles CO.

This is approximately 22.4L /mol * 0.232 moles = 5.20 L at “STP” - slightly more a “NTP”.
This would produce the same molar and volumetric amount of $CO_2$.

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