1, g of a carbonate (M_{2}CO_{3}) on treatmen | vedclass

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(B) The balanced chemical equation for the reaction is:$M_{2}CO_{3} + 2HCl \rightarrow 2MCl + H_{2}O + CO_{2}$According to the stoichiometry of the re

Vedclass · Accepted Answer

$84.3$

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(B) The balanced chemical equation for the reaction is:$M_{2}CO_{3} + 2HCl ightarrow 2MCl + H_{2}O + CO_{2}$According to the stoichiometry of the reaction,$1\, mol$ of $M_{2}CO_{3}$ produces $1\, mol$ of $CO_{2}$.Therefore,the moles of $M_{2}CO_{3}$ reacted is equal to the moles of $CO_{2}$ produced,which is $0.01186\, mol$.Using the formula: $	ext{Molar mass} = \frac{	ext{Mass}}{	ext{Moles}}$$	ext{Molar mass} = \frac{1\, g}{0.01186\, mol} \approx 84.3\, g\, mol^{-1}$.

$1\, g$ of a carbonate $(M_{2}CO_{3})$ on treatment with excess $HCl$ produces $0.01186\, mol$ of $CO_{2}.$ The molar mass of $M_{2}CO_{3}$ in $g\, mol^{-1}$ is:

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