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The mass of $\mathrm{BaCO}_3$ produced when excess $\mathrm{CO}_2$ is bubbled through a solution of $0.205 \mathrm{~mol}$ $\mathrm{Ba}(\mathrm{OH})_2$ is
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$40.5 \mathrm{~g}$
$\mathrm{Ba}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{BaCO}_3+\mathrm{H}_2 \mathrm{O}$
Atomic wt. of $\mathrm{BaCO}_3=137+12+16 \times 3=197$
No. of mole $=\frac{\text { wt. of substance }}{\text { mol wt. }}$
囚 1 mole of $\mathrm{Ba}(\mathrm{OH})_2$ gives 1 mole of $\mathrm{BaCO}_3$
$\therefore 205$ mole of $\mathrm{Ba}(\mathrm{OH})_2$ will give 205 mole of $\mathrm{BaCO}_3$
$\therefore$ wt. of 0.205 mole of $\mathrm{BaCO}_3$ will be $.205 \times 197=40.385 \mathrm{gm} \approx 40.5 \mathrm{gm}$
Atomic wt. of $\mathrm{BaCO}_3=137+12+16 \times 3=197$
No. of mole $=\frac{\text { wt. of substance }}{\text { mol wt. }}$
囚 1 mole of $\mathrm{Ba}(\mathrm{OH})_2$ gives 1 mole of $\mathrm{BaCO}_3$
$\therefore 205$ mole of $\mathrm{Ba}(\mathrm{OH})_2$ will give 205 mole of $\mathrm{BaCO}_3$
$\therefore$ wt. of 0.205 mole of $\mathrm{BaCO}_3$ will be $.205 \times 197=40.385 \mathrm{gm} \approx 40.5 \mathrm{gm}$
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