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(B) The balanced chemical equation is: $Ba(OH)_2 + CO_2 \rightarrow BaCO_3 + H_2O$ From the stoichiometry,$1 \ mol$ of $Ba(OH)_2$ produces $1 \ mol$ o

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$40.5$

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(B) The balanced chemical equation is: $Ba(OH)_2 + CO_2 ightarrow BaCO_3 + H_2O$ From the stoichiometry,$1 \ mol$ of $Ba(OH)_2$ produces $1 \ mol$ of $BaCO_3$. Therefore,$0.205 \ mol$ of $Ba(OH)_2$ will produce $0.205 \ mol$ of $BaCO_3$. The molar mass of $BaCO_3 = 137 + 12 + (3 	imes 16) = 197 \ g \ mol^{-1}$. Mass of $BaCO_3 = 	ext{moles} 	imes 	ext{molar mass} = 0.205 \ mol 	imes 197 \ g \ mol^{-1} = 40.385 \ g \approx 40.5 \ g$ (rounding to the nearest option).

How many grams of $BaCO_3$ are produced when excess $CO_2$ is passed through $0.205 \ mol$ of $Ba(OH)_2$?

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