Question

$\text{C}{{\text{H}}_{4}}+\text{2}{{\text{O}}_{2}}\to \text{C}{{\text{O}}_{2}}+\text{2}{{\text{H}}_{2}}\text{O}$
As per this reaction, if 20 mL of $\text{C}{{\text{H}}_{4}}$ and 20 mL of ${{\text{O}}_{2}}$ were exploded together and if the reaction mixture is cooled to room temp, the resulting gas mixture will have a volume of:
A. 40 mL
B. 20 mL
C. 60 mL
D. 30 mL

Answer 1

To solve this question, solve the numerical according to stoichiometric coefficients. Check the temperature of the reaction. The concept used here is limiting reagent. Check limiting reagent by calculating the volume of a substance required by the help of the volume of another substance. Then, find the volume of carbon dioxide and volume of leftover substance.

Complete step by step answer:
Let us discuss the numerical from the basics and through root steps:
According to stoichiometric coefficients, 1 mole of methane $\left( \text{C}{{\text{H}}_{4}} \right)$ reacts with 2 moles of oxygen gas $\left( {{\text{O}}_{2}} \right)$ to form 1 mole of carbon dioxide gas $\left( \text{C}{{\text{O}}_{2}} \right)$ and 1 mole of liquid water $\left( {{\text{H}}_{2}}\text{O} \right)$.
When a reaction involving gaseous substances occurs at the same temperature and pressure, the mole ratios are treated the same as the ratio between their volumes. So, 1 mL of methane $\left( \text{C}{{\text{H}}_{4}} \right)$ reacts with 2 mL of oxygen gas $\left( {{\text{O}}_{2}} \right)$ to form 1 mL of carbon dioxide gas $\left( \text{C}{{\text{O}}_{2}} \right)$.
Now, here, the concept to be used is ‘limiting reagent’. A substance which gets totally consumed when the reaction gets completed. The amount of product is formed according to limiting reagent.
Let us find the limiting reagent for this reaction, 1 mL of methane $\left( \text{C}{{\text{H}}_{4}} \right)$ reacts with 2 mL of oxygen gas $\left( {{\text{O}}_{2}} \right)$.
Then, 20 mL of methane $\left( \text{C}{{\text{H}}_{4}} \right)$ will need 40 mL of oxygen gas $\left( {{\text{O}}_{2}} \right)$.
The amount of oxygen gas $\left( {{\text{O}}_{2}} \right)$ given in the question is 20 mL.
So, it is clear that oxygen gas $\left( {{\text{O}}_{2}} \right)$ will be limiting reagent. As, the requirement is 40 mL but present is only 20 mL. It will finish early in the reaction. The products will be formed according to oxygen gas $\left( {{\text{O}}_{2}} \right)$.
2 mL of oxygen gas $\left( {{\text{O}}_{2}} \right)$ to form 1 mL of carbon dioxide gas $\left( \text{C}{{\text{O}}_{2}} \right)$. 20 mL of oxygen gas will form $\dfrac{20}{2}$ mL volume of carbon dioxide $\left( \text{C}{{\text{O}}_{2}} \right)$.
The volume of carbon dioxide $\left( \text{C}{{\text{O}}_{2}} \right)$ is 10 mL.
1 mL of methane $\left( \text{C}{{\text{H}}_{4}} \right)$ reacts with 2 mL of oxygen gas $\left( {{\text{O}}_{2}} \right)$. 20 mL of oxygen gas uses $\dfrac{20}{2}$ mL of the volume of methane $\left( \text{C}{{\text{H}}_{4}} \right)$.
The amount of methane gas $\left( \text{C}{{\text{H}}_{4}} \right)$ used is 10 mL and volume of methane gas $\left( \text{C}{{\text{H}}_{4}} \right)$ left is 20 mL – 10 mL or 10 mL.
- The total mixture of volumes of gases is 20 mL; 10 mL from carbon dioxide gas $\left( \text{C}{{\text{O}}_{2}} \right)$ and 10 mL from methane gas $\left( \text{C}{{\text{H}}_{4}} \right)$.

The correct option is option ‘b’.

Note: The question arises while solving this numerical is why are we considering water as liquid not as gaseous vapours? The answer to this question is the temperature at which reaction is taking place. The temperature of the reaction is ${{25}^{\text{o}}}\text{C}$. At this temperature, water exists in a liquid state not vapour state. Water exists in vapours above ${{100}^{\text{o}}}\text{C}$.