Question

The work done in an open vessel at $300K$, when 112 g iron reacts with $HCl$ to give $FeC{l}_2$, is nearly:

• A. $-1.2kcal$
• B. $0.6kcal$
• C. $-0.3kcal$
• D. $-0.2kcal$

Answer 1

Answer: (A)

$-1.2kcal$

Solution:-

Molar mass of $Fe=56g$

Given mass of $Fe=112g$

As we know that,

$\text{No. of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Therefore, number of moles of $Fe$ used-

$n=\frac{112}{56}=2\text{moles}$

As we know that,

$w=-P\Delta V=-nRT[\because PV=nRT]$

Given $T=300K$

$\therefore w=-2\times 2\times 300=-1200\text{cal}=-1.2kcal$

Hence the work done will be $-1.2\text{kcal}$.