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1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2.  The molar mass of M2CO3 in g mol−1 is :

Answers (1)

Number of Moles -

No of moles = given mass of substance/ molar mass of substance

Given Chemical reaction,

M_2CO_3 + 2HCl \rightarrow 2MCl + 2H_2O + CO_2

    1gm                                                           0.01186\;moles

From the balanced equation 

\frac{1}{M} = 0.01186 \Rightarrow M = \frac{1}{0.01186} = 83.4

Posted by

Satyajeet Kumar

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