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

Engineering

Chemistry

Mixture Related Problems

Question

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:

11.86

84.3

1186

118.6

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M₂CO₃ + 2HCl $\to_{}^{}$ 2MCl + H₂O + CO₂

_{$n_{M_{2} {CO}_{3}} = n_{{CO}_{2}} = 0 .01186$}

0.01186 mol $\to_{}^{}$ 1 gm

$1 mo \overset{}{\to} (\frac{1}{0 .01186} \times 1)$

= 84.3 g mol^–1