Question

Calculate the work done when 112 g iron reacts with dilute HCl at 300 K. The reaction is carried out in an open container

(At mass of Fe = 56)

• A. 600 cal
• B. 300 cal
• C. 200 cal
• D. 1200 cal

Answer 1

Answer: (D)

The number of moles of Fe taken

$=\frac{112}{56}=2$

So the chemical reaction is

$2Fe\left(s\right)+4HCl\left(a q\right) \rightarrow 2FeCl_{2}\left(a q\right)+2H_{2}\left(g\right)$

New work done $=P\left(V_{2} - V_{1}\right)$

Since $V_{2}-V_{1}=V_{H_{2}}$

Hence volume of solids and liquids is taken as $\cong0$

So work done $=P.V_{H_{2}}$

also $PV_{H_{2}}=nRT$ or $V_{H_{2}}=\frac{n R T}{P}$

So work done $=P.\frac{n R T}{P}=nRT$ [ $R=2$ Cal]

$=2\times 2\times 300=1200$ Cal.