Question

What will be the volume of the mixture after the reaction?

$\mathbf{PV = nRT}$

• A. 0.5 litre
• B. 1 litre
• C. 2.5 litre
• D. 0.1 litre

Answer 1

Answer: (C)

2.5 litre

Answer 2

$\frac{2p_{1}}{T_{1} + T_{2}}$

$H_{2}$

$\mathbf{= 6.02 \times 1}\mathbf{0}^{\mathbf{21}}$ is a limiting compound. That's why $(\bar{v})$ of $\bar{v} = 0.9213u$ reacts with $(3RT/M)^{1/2}$ of $V_{m} > 22.4$ and forms$V_{m} < 22.4$.

Thus (4 - 1.5) $V_{m} = 22.4$ $V_{m} = 44.8$ remains after the reaction.