Question:
A reaction gas mixture contains \(50\%\), \(30\%\), \(20\%\) of \(A\), \(B\) and \(C\) by volume, respectively. The mixture undergoes the following reactions.
\[
1.\quad A+2B \longrightarrow P_1
\]
\[
2.\quad 4P_1+3C \longrightarrow P_2
\]
\(P_1\) and \(P_2\) are two products of the reactions. Choose the correct answer if the reaction completes.
A reaction gas mixture contains \(50\%\), \(30\%\), \(20\%\) of \(A\), \(B\) and \(C\) by volume, respectively. The mixture undergoes the following reactions.
\[ 1.\quad A+2B \longrightarrow P_1 \] \[ 2.\quad 4P_1+3C \longrightarrow P_2 \] \(P_1\) and \(P_2\) are two products of the reactions. Choose the correct answer if the reaction completes.
\[ 1.\quad A+2B \longrightarrow P_1 \] \[ 2.\quad 4P_1+3C \longrightarrow P_2 \] \(P_1\) and \(P_2\) are two products of the reactions. Choose the correct answer if the reaction completes.
Show Hint
For gaseous reactions, volume ratios follow the stoichiometric coefficients when temperature and pressure are the same. First find the limiting reactant in each reaction, then decide which species is completely exhausted.
Updated On: Jun 26, 2026
- \(C\) will be completely exhausted.
- \(A\) will be completely exhausted.
- \(B\) will not be completely exhausted.
- \(P_1\) will be completely exhausted.
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Assume total volume of gaseous mixture.
Let the total volume of the mixture be \[ 100 \] So, the volumes of \(A\), \(B\), and \(C\) are \[ A=50,\quad B=30,\quad C=20 \]
Step 2: Use the first reaction.
The first reaction is \[ A+2B \longrightarrow P_1 \] According to the reaction, \[ 1 \text{ volume of } A \] reacts with \[ 2 \text{ volumes of } B \] Since \(B=30\), the amount of \(A\) required is \[ \frac{30}{2}=15 \] So, \(B\) is completely used and \(15\) volumes of \(A\) are used.
Thus, \(P_1\) formed is \[ P_1=15 \] Remaining \(A\) is \[ 50-15=35 \] Remaining \(C\) is still \[ 20 \]
Step 3: Use the second reaction.
The second reaction is \[ 4P_1+3C \longrightarrow P_2 \] For \(4\) volumes of \(P_1\), required \(C\) is \(3\) volumes.
Therefore, for \(15\) volumes of \(P_1\), required \(C\) is \[ \frac{3}{4}\times 15 \] \[ =\frac{45}{4} \] \[ =11.25 \]
Step 4: Compare available \(C\) with required \(C\).
Available \(C\) is \[ 20 \] Required \(C\) is \[ 11.25 \] Since \[ 20\gt 11.25, \] \(C\) is not completely exhausted.
But all \(15\) volumes of \(P_1\) are consumed in the second reaction.
Therefore, \[ P_1 \] will be completely exhausted.
Step 5: Final conclusion.
Hence, the correct statement is \[ \boxed{P_1\text{ will be completely exhausted.}} \]
Let the total volume of the mixture be \[ 100 \] So, the volumes of \(A\), \(B\), and \(C\) are \[ A=50,\quad B=30,\quad C=20 \]
Step 2: Use the first reaction.
The first reaction is \[ A+2B \longrightarrow P_1 \] According to the reaction, \[ 1 \text{ volume of } A \] reacts with \[ 2 \text{ volumes of } B \] Since \(B=30\), the amount of \(A\) required is \[ \frac{30}{2}=15 \] So, \(B\) is completely used and \(15\) volumes of \(A\) are used.
Thus, \(P_1\) formed is \[ P_1=15 \] Remaining \(A\) is \[ 50-15=35 \] Remaining \(C\) is still \[ 20 \]
Step 3: Use the second reaction.
The second reaction is \[ 4P_1+3C \longrightarrow P_2 \] For \(4\) volumes of \(P_1\), required \(C\) is \(3\) volumes.
Therefore, for \(15\) volumes of \(P_1\), required \(C\) is \[ \frac{3}{4}\times 15 \] \[ =\frac{45}{4} \] \[ =11.25 \]
Step 4: Compare available \(C\) with required \(C\).
Available \(C\) is \[ 20 \] Required \(C\) is \[ 11.25 \] Since \[ 20\gt 11.25, \] \(C\) is not completely exhausted.
But all \(15\) volumes of \(P_1\) are consumed in the second reaction.
Therefore, \[ P_1 \] will be completely exhausted.
Step 5: Final conclusion.
Hence, the correct statement is \[ \boxed{P_1\text{ will be completely exhausted.}} \]
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