MCQ

Limiting Reagent and Yield

By Sakshi Goel | 19 May'20 | 1 K Views

MCQ

Problem:
The mass of BaCO₃ produced when excess CO₂ is bubbled through a solution of 0.205 mol Ba(OH)₂ is
(a) 81 g
(b) 20.25 g
(c) 40.5 g
(d) 162 g

Solution:

The reaction is:
$Ba (OH)_{2} + {CO}_{2} â†’ {BaCO}_{3} + H_{2} O$
CO₂ is in excess, so Ba(OH)₂ is the limiting reagent.
Moles of Ba(OH)₂reacted = 0.205 mol
Moles of BaCO₃ formed = Moles of Ba(OH)₂reacted = 0.205 mol
Molar mass of BaCO₃ = 197 g/mol
Mass of BaCO₃ formed = 0.205 x 197 = 40.4 g
Closest option is (c).