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

$2Fe\left(s\right)+4HCl\left(aq\right)\to 2FeC{l}_2\left(aq\right)+2H_2\left(g\right)$

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 $\cong 0$

So work done $=P.V_{H_2}$

also $P{V}_{H_2}=nRT$ or $V_{H_2}=\frac{nRT}{P}$

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

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