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Stoichiometric Calculations

1 gram of a c...

Question

1 gram of a carbonate $(M_{2} C O_{3})$ on treatment with excess $H C l$ produces 0.01186 moles of $C O_{2}$ . The molar mass of $(M_{2} C O_{3})$ in $g m o l^{- 1}$ is:

84.3

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118.6

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11.86

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1186

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Solution

The correct option is A 84.3
$M_{2} C O_{3} + 2 H C l \to C O_{2} + 2 M C l + H_{2} O$

Moles of $M_{2} C O_{3}$ = Moles of $C O_{2}$ produced.

Moles of $M_{2} C O_{3} = \frac{w}{m o l a r m a s s} = 0.01186$

Weight of the carbonate $(w)$ = 1 $g$

Moles of $M_{2} C O_{3} = \frac{1}{m o l a r m a s s} = 0.01186$

$∴$ Molar mass $= 84.3 g m o l^{- 1}$

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