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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 $0.01186$ mole 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
$M_{2} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + C O_{2}$
$0.01186$ moles $C O_{2} =$ $0.01186$ moles of $M_{2} C O_{3} =$ $1 g$ $M_{2} C O_{3}$
Molar mass of $M_{2} C O_{3} = \frac{M a s s o f M_{2} C O_{3}}{N o . o f m o l e s o f M_{2} C O_{3}} = \frac{1 g}{0.01186 m o l e s} = 84.3 g / m o l$

Hence, the correct option is $A$ .

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