MCQ

Limiting Reagent and Yield

By Sakshi Goel | 19 May'20 | 928 Views

MCQ

Problem:
2.76 g of silver carbonate on being strongly heated at quite high temperature yields a residue weighing
(a) 2.64 g
(b) 2.48 g
(c) 2.32 g
(d) 2.16 g

Solution:

The reaction of heating silver carbonate is:
${Ag}_{2} {CO}_{3} â†’ 2 Ag (s) + {CO}_{2} (g) + \frac{1}{2} O_{2} (g)$
In the reaction, only Ag is residue.
Molar mass of Ag₂CO₃ = 276 g/mol
Moles of Ag₂CO₃ in 2.76 g = mass/molar mass = 2.76/276 = 0.01 mol
1 mole of Ag₂CO₃ gives 2 moles of Ag
So, 0.01 mol of Ag₂CO₃ gives 0.02 moles of Ag
Molar mass of Ag = 108 g/mol
Mass of 0.02 mol of Ag = 108 x 0.02 = 2.16 g
Mass of residue = 2.16 g
Option (d) is correct.