The mass of BaCO3 produced, when excess CO2 is bubbled through a solution of 0.205 moles of Ba(OH)2 is

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The mass of BaCO₃ produced, when excess CO₂ is bubbled through a solution of 0.205 moles of Ba(OH)₂ is

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40.5 g

20.25 g

162 g

81 g

answer is B.

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Detailed Solution

$Ba (OH)_{2} + {CO}_{2} \to {BaCO}_{3} + H_{2} O$
$Atomic wt. of {BaCO}_{3} = 137 + 12 + 16 \times 3 = 197$
$No. of mole = \frac{wt. of substance}{mol wt.}$
$∵ 1 mole of Ba (OH)_{2} gives 1 mole of {BaCO}_{3}$
$∴ 0.205 mole of Ba (OH)_{2} will give 0.205 {mole of BaCO}_{3}$
$∴ wt. of 0.205 mole of {BaCO}_{3} will be$
$0.205 \times 197 = 40.385 gm \approx 40.5 gm$