Question

One gram of a carbonate $\left( {{M_2}C{O_3}} \right)$on treatment with excess $HCl$ produces $0.01186$ moles of $C{O_2}$. The molar mass of ${M_2}C{O_3}$in $gm{\left( {mol} \right)^{ - 1}}$
A.$84.3$
B.$118.6$
C.$11.86$
D.$1186$

Answer 1

The molar mass of the compounds can be determined from the number of the moles of the compound and weight of the compound. The number of moles can be determined from the chemical reaction involved between the reactants and products.
Formula used:
$n = \frac{w}{M}$
n is number of moles of ${M_2}C{O_3}$
w is the weight of ${M_2}C{O_3}$ in grams
M is the molar mass of ${M_2}C{O_3}$ in grams

Complete answer:
Metals are involved in chemical bonding with carbonate and form metal carbonates. The metal carbonate in the given problem has the formula of ${M_2}C{O_3}$.
Metal carbonates when treated with hydrochloric acid forms metal chlorides i.e.. salts along with the liberation of water and carbon dioxide.
The chemical reaction will be as follows:
${M_2}C{O_3} + 2HCl \to 2MCl + {H_2}O + C{O_2}$
From the above chemical reaction, the number of moles of carbon dioxide is equal to number of moles of ${M_2}C{O_3}$
The number of moles of carbon dioxide given is $0.01186$moles
Thus, number of moles of ${M_2}C{O_3}$ will be $0.01186$moles
The given weight of ${M_2}C{O_3}$ is $1g$
By substituting the values of number of moles and weight of carbonate in the formula, we will get
$M = \frac{w}{n} = \frac{1}{{0.01186}} = 84.3gm{\left( {mol} \right)^{ - 1}}$
The molar mass of ${M_2}C{O_3}$ in $gm{\left( {mol} \right)^{ - 1}}$ is $84.3gm{\left( {mol} \right)^{ - 1}}$
Thus option A is the correct answer.

Note:
The number of moles of reactants and products can be clearly known from the chemical reaction. As the number of moles of carbon dioxide and metal carbonate are equal. If the number of moles of products are double to the moles of reactants, then it should be multiplied by two.