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  4. The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is

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Q. The mass of $BaCO_3$ produced when excess $CO_2$ is bubbled through a solution of $0.205 \,mol\, Ba\left(OH\right)_2$ is

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Solution:

$Ba\left(OH\right)_2 + CO_2 \to Ba \,CO_3 + H_2O$
Atomic $wt. of BaCO_3 = 137 + 12 + 16 \times 3 = 197$
$No. of mole =\frac{wt. of substance}{mol wt}$
$\because \,1 \,mole$ of $Ba\left(OH\right)_2$ gives $1$ mole of $BaCO_3$
$\therefore 0.205 moles of Ba\left(OH\right)_2$ will give $0.205 mole\, of\, BaCO_3$
$\therefore wt. of 0.205$ mole of $BaCO_3$ will be
$0.205 \times 197= 40.385 g\approx 40.5 g$