Question

The work is done in an open vessel when $113$ gm of $Fe$ reacts with dil. $HCl$ to form $FeC{l}_2$ at ${27}^o$C will be:

• A. $w=-1200$ cal
• B. $w=-600$ cal
• C. $w=-300$ cal
• D. $w=-200$ cal

Answer 1

Answer: (A)

$w=-1200$ cal

Solution:- (A) $w=-1200\text{cal}$

Molar mass of $Fe=56g$

Given mass of $Fe=113g$

As we know that,

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

Therefore, number of moles of $Fe$ used-

$n=\frac{113}{56}=2.018\text{moles}$

As we know that,

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

Given $T=27℃=(27+273)=300K$

$\therefore w=-2.018\times 2\times 300=1210.8$

$\approx -1200\text{cal}$

Hence the work done will be $-1200\text{cal}$.