The mass of $\mathrm{BaCO}_{3}$ produced when excess of $\mathrm{CO}_{2}$ is bubbled through a solution of…

The mass of $\mathrm{BaCO}_{3}$ produced when excess of $\mathrm{CO}_{2}$ is bubbled through a solution of $0.205 \mathrm{~mol} \mathrm{Ba}(\mathrm{OH})_{2}$ is
  1. $81 \mathrm{~g}$
  2. $40.5 \mathrm{~g}$
  3. $20.25 \mathrm{~g}$
  4. $162 \mathrm{~g}$

Solution

$\mathrm{Ba}(\mathrm{OH})_{2}+\mathrm{CO}_{2} ightarrow \mathrm{BaCO}_{3}+\mathrm{H}_{2} \mathrm{O}$
Atomic wt. of $B a C O_{3}=137+12+16 \times 3=197$
No. of mole $=\frac{w t \text { of substance }}{\text { mol wt. }}$
$\because$ 1 mole of $\mathrm{Ba}(\mathrm{OH})_{2}$ gives 1 mole of $\mathrm{BaCO}_{3}$
$\therefore$ 205 mole of $B a(O H)_{2}$ will give 205 mole of $B a C O_{3}$
$\therefore$ wt. of $0.205$ mole of $\mathrm{BaCO}_{3}$ will be
$0.205 \times 197=40.385 \mathrm{gm}=40.5 \mathrm{gm}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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