Q. The mass of ${BaCO}_{3}$ produced when excess ${CO}_{2}$ is bubbled through a solution of 0.205 mole Ba$(OH)_{2}$ is,
A) 81 g
B) 40.5 g
C) 20.25 g
D) 162 g
Explanation
(b)
Given, moles of $Ba(OH)_{2}$=0.205
$Ba(OH)_{2} + CO_{2} \rightarrow BaCO_{3} + H_{2}O$
$\therefore$ 0.205 moles of $Ba(OH)_{2} \equiv$0.205 moles of $BaCO_{3}$
$\therefore$ Mass of $BaCO_{3}$=moles of $BaCO_{3} \times$ molecules mass of $BaCO_{3}$
$= 0.205 \times 197.3$
$= 40.5$ g
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.