Calculate the work done in an open vessel at 300 K, when $168 g$ iron having molar mass $56 g / m o l$ reacts with $d i l . H C l$
$R \approx 2 c a l / m o l K$

1200 c a l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1400 c a l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1800 c a l

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2400 c a l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1800 c a l$
$F e + 2 H C l \to F e C l_{2} + H_{2} 1 m o l 1 m o l \frac{168}{56} = 3 m o l 3 m o l N o w, V_{2} = V_{H_{2}} a n d V_{1} = 0$
(For solid + liquid state)
$Work done (W) = p (v_{2} - v_{1}) W = p (H_{2} - 0) W = p V_{H_{2}} W = n_{H_{2}} R T W = 3 \times 2 \times 300 W = 1800 c a l$

Suggest Corrections

Similar questions

56 g

H_{2} S O_{4}

30^{o} C

(I)

(I I)

F e

56 g/mol

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program