When CO_2 is bubbled through 0.205 mol of Ba | vedclass

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Question

(B) The balanced chemical equation is: $Ba(OH)_2 + CO_2 ightarrow BaCO_3 + H_2O$ The molar mass of $BaCO_3 = 137 + 12 + (16 	imes 3) = 197 \ g/mol$

Vedclass · Accepted Answer

$40.5$

Vedclass · Answer

(B) The balanced chemical equation is: $Ba(OH)_2 + CO_2 ightarrow BaCO_3 + H_2O$ The molar mass of $BaCO_3 = 137 + 12 + (16 	imes 3) = 197 \ g/mol$. According to 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 mass of $BaCO_3$ produced $= 0.205 \ mol 	imes 197 \ g/mol = 40.385 \ g$. Rounding to the nearest provided option,the answer is $40.5 \ g$.

When $CO_2$ is bubbled through $0.205 \ mol$ of $Ba(OH)_2$,how many grams of $BaCO_3$ will be produced?

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