Calculate the composition of the resulting mixture when $450 mL$ of carbon monoxide is mixed with $200 mL$ of oxygen.

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Solution

Step1. Given information
$450 mL$ of carbon monoxide $(CO)$ , $200 mL$ of oxygen $(O_{2})$ .
Step2. Application of Gay-Lussac's Law of combining volumes
According to this law, when gases react, the ratio of the volume of gaseous reactant to the gaseous product is expressed as a simple whole-number ratio. The conditions required are constant pressure and temperature.
This law is only applicable to gases. As a result, the volumes of solids and liquids are assumed to be zero.
For the given information, it is applied as follows:
$2 CO + O_{2} \to 2 {CO}_{2} 2 vols 1 vols 2 vols$
Step3. Calculation of composition of the resulting mixture
$1 vol$ of oxygen react with $2 vol$ of carbon monoxide
$200 mL$ of oxygen will react with $200 mL \times 2 = 400 mL$ of carbon monoxide.
Remaining carbon monoxide
$= 450 mL-400 mL =50 mL$
$1 vol$ of oxygen produces $2 vol$ of carbon dioxide
$200 mL$ of oxygen produces $200 mL \times 2 = 400 mL$ of carbon dioxide.
Hence, the composition of the resulting mixture is $400 mL$ of carbon dioxide, and $50 mL$ of carbon monoxide.

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Calculate the composition of the resulting mixture when 450mL of carbon monoxide is mixed with 200mL of oxygen.

Calculate the composition of the resulting mixture when $450 mL$ of carbon monoxide is mixed with $200 mL$ of oxygen.