MCQ

Limiting Reagent and Yield

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

MCQ

Problem:
Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g BaCO₃ will be
(a) 1.12 L
(b) 0.84 L
(c) 2.24 L
(d) 4.06 L

Solution:

The reaction is:
${BaCO}_{3} â†’ BaO + {CO}_{2}$
molar mass of BaCO₃ = 197 g/mol
Mass of BaCO₃ decomposed = 9.85 g
Moles of BaCO₃ decomposed = 9.85/197 = 0.05 mol
1 mole of BaCO₃ decomposes to give 1 mole of CO₂.
0.05 mole of BaCO₃ decomposes to give 0.05 mole of CO₂.
At STP, Volume of 1 mole of gas = 22.4 L
Volume of 0.05 mol of gas = 22.4 x 0.05 = 1.12 L
Correct option is (a).