Question

${\text{10}}$ liter of ${{\text{O}}_{\text{2}}}$ gas is reacted with ${\text{30}}$ liter of ${\text{CO}}$ gas at STP. The volumes of each gas present at the end of reaction are:
A. ${\text{CO}}$ (${\text{10}}$ liter), ${\text{C}}{{\text{O}}_{\text{2}}}$ (${\text{20}}$ liter)
B. ${\text{10}}$ ${{\text{O}}_{\text{2}}}$ (${\text{5}}$ liter), ${\text{C}}{{\text{O}}_{\text{2}}}$ (${\text{20}}$ liter)
C. ${\text{CO}}$ (${\text{20}}$ liter), ${\text{C}}{{\text{O}}_{\text{2}}}$ (${\text{10}}$ liter)
D. ${{\text{O}}_{\text{2}}}$ (${\text{10}}$ liter), ${\text{C}}{{\text{O}}_{\text{2}}}$ (${\text{20}}$ liter)
E. ${{\text{O}}_{\text{2}}}$ ( liter), ${\text{CO}}$ (${\text{10}}$ liter)

Answer 1

The reaction of ${{\text{O}}_{\text{2}}}$ gas with ${\text{CO}}$ gas produces ${\text{C}}{{\text{O}}_2}$ gas. Initially, write the correct balanced reaction. Determine the stoichiometric amounts of the reactants and products to calculate the volume of each gas present at the end of the reaction.

Complete step by step answer:
Step 1:
The reaction of ${{\text{O}}_{\text{2}}}$ gas with ${\text{CO}}$ gas produces ${\text{C}}{{\text{O}}_2}$ gas. Thus, the reaction is,
${{\text{O}}_{\text{2}}} + {\text{CO}} \to {\text{C}}{{\text{O}}_2}$
Change the coefficient of ${\text{CO}}$ to $2$ to balance the number of oxygen atoms. Thus,
${{\text{O}}_{\text{2}}} + 2{\text{CO}} \to {\text{C}}{{\text{O}}_2}$
Change the coefficient of ${\text{C}}{{\text{O}}_2}$ to $2$ to balance the number of carbon atoms. Thus, the balanced reaction is,
${{\text{O}}_{\text{2}}} + 2{\text{CO}} \to 2{\text{C}}{{\text{O}}_2}$
Step 2:
From the balanced reaction,
$1$ volume of ${{\text{O}}_{\text{2}}}$ gas reacts with $2$ volumes of ${\text{CO}}$ to produce $2$ volumes of ${\text{C}}{{\text{O}}_2}$ gas.
We have ${\text{10}}$ liter of ${{\text{O}}_{\text{2}}}$ gas. Thus stoichiometrically,
${\text{10}}$ liters of ${{\text{O}}_{\text{2}}}$ gas react with ${\text{20}}$ liters of ${\text{CO}}$ gas to produce ${\text{20}}$ liters of ${\text{C}}{{\text{O}}_2}$ gas.
Step 3:
At the beginning of the reaction, we have, ${\text{10}}$ liter of ${{\text{O}}_{\text{2}}}$ gas and ${\text{30}}$ liter of ${\text{CO}}$ gas.
Thus, after the reaction,
${\text{CO}}$ remaining $= \left( {30 - 20} \right) = 10$ liters.
${{\text{O}}_2}$ remaining $= \left( {10 - 10} \right) = 00$ liters.
${\text{C}}{{\text{O}}_2}$ produced $= 20$ liters.
Thus, at the end of the reaction, the volume of gases are ${\text{CO}}$ (${\text{10}}$ liter), ${{\text{O}}_{\text{2}}}$ (${\text{00}}$ liter) and ${\text{C}}{{\text{O}}_{\text{2}}}$ (${\text{20}}$ liter).
So, the correct answer is “Option A”.

Note: Balancing the reaction to get the correct volumes of gases is important. An unbalanced reaction can lead to incorrect volumes of gases.