1 ~g of a carbonate (M_2 CO_3) on treatment with excess HCl produces 0.01 ~mol of CO_2. The molar mass of M_2 CO_3 is __________ g ~mol^-1. (Nearest integer)

When the carbonate reacts with HCl, it produces CO 2 , as shown in the following balanced equation: M_2CO_3 + 2HCl 2MCl + H_2O + CO_2 From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO 2 . To determine the molar mass of the carbonate, we can use the stoichiometry of the reac

1 g of a carbonate M2 CO3 on treatment with excess HCl produ: null (Chemistry)

When the carbonate reacts with HCl, it produces CO₂, as shown in the following balanced equation:

\mathrm{M}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{MCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2

From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO₂.

To determine the molar mass of the carbonate, we can use the stoichiometry of the reaction:

1 mol of

\mathrm{M}_2\mathrm{CO}_3

produces 1 mol of CO2.

So, 0.01 mol of CO₂ corresponds to 0.01 mol of

\mathrm{M}_2\mathrm{CO}_3

.

Now, we can find the molar mass of

\mathrm{M}_2\mathrm{CO}_3

\frac{\text{mass of }\mathrm{M}_2\mathrm{CO}_3}{\text{moles of }\mathrm{M}_2\mathrm{CO}_3} = \text{molar mass of }\mathrm{M}_2\mathrm{CO}_3

\frac{1 \text{ g}}{0.01 \text{ mol}} = 100 \text{ g mol}^{-1}

So, the molar mass of

\mathrm{M}_2\mathrm{CO}_3

is approximately 100 g/mol.

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