One gram of a metal carbonate (M_ 2 CO_ 3 ) reacts with an excess of HCl to produce 0.01186 text moles of CO_ 2 . Calculate the molar mass of M_ 2 CO_ 3 in text g mol ^ -1 .

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EEJ MAIN Chemistry QUESTION #7186

Question 1

One gram of a metal carbonate ($M_{2}CO_{3}$) reacts with an excess of $HCl$ to produce $0.01186 \text{ moles}$ of $CO_{2}$. Calculate the molar mass of $M_{2}CO_{3}$ in $\text{g mol}^{-1}$.

$84.3$✔️
$118.6$
$11.86$
$1186$

Correct Answer Explanation

The reaction is $M_{2}CO_{3} + 2HCl \rightarrow 2MCl + H_{2}O + CO_{2}$:

From stoichiometry, $1 \text{ mole}$ of $M_{2}CO_{3}$ produces $1 \text{ mole}$ of $CO_{2}$.
Moles of $M_{2}CO_{3} = \text{Moles of } CO_{2} = 0.01186$.
Molar Mass = $\text{Mass} / \text{Moles} = 1 \text{ g} / 0.01186 \approx 84.3 \text{ g/mol}$.

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