Animated Solution for Chemistry - Basic Concepts in Chemistry: 1 g of a carbonate ($M_{2}C{O}_3$) on treatment with excess HCl produces 0.01186 mole of $C{O}_2$. The molar mass of $M_{2}C{O}_3$ in ${\text{g mol}}^{-1}$ is

Imagine you are standing in a chemistry laboratory. In front of you is a beaker filled with a clear liquid—hydrochloric acid ($HCl$). You are handed exactly $1\text{ g}$ of a mysterious white powder, a metal carbonate with the general formula $M_{2}C{O}_3$.

The moment you drop this powder into the acid, a vigorous reaction begins. The liquid fizzes and bubbles violently. This effervescence is the visual signature of carbon dioxide ($C{O}_2$) gas escaping into the atmosphere. The problem tells us that exactly $0.01186\text{ moles}$ of this gas were produced. Our mission? To play chemical detective and find the molar mass of the original mystery powder.

In chemistry, the balanced equation is our ultimate map. It tells us exactly how the atoms rearrange themselves. Let's write down the reaction between our metal carbonate and hydrochloric acid:

Notice the stoichiometry here. The coefficients in front of the molecules reveal their molar ratios. For every one mole of $M_{2}C{O}_3$ that reacts, exactly one mole of $C{O}_2$ gas is produced. The $HCl$ is in excess, which means our metal carbonate is the limiting reagent. It dictates exactly how much product can be formed.

Because the molar ratio of $M_{2}C{O}_3$ to $C{O}_2$ is $1:1$, we can establish a beautiful, simple mathematical bridge:

We know that the number of moles of any substance is its given mass divided by its molar mass. Let's denote the unknown molar mass of our carbonate as $M$.

Therefore, the moles of the carbonate can be written as:

We are already given the moles of $C{O}_2$ produced:

Now, we simply equate the two expressions based on our $1:1$ stoichiometric ratio:

To find $M$, we just need to take the reciprocal of $0.01186$. Don't let the decimals intimidate you.

Looking at our options, 84.3 is the perfect match.

Who is the Mystery Metal?

We found the molar mass of the entire compound $M_{2}C{O}_3$ to be $84.3{\text{ g mol}}^{-1}$. But what if we wanted to identify the metal $M$ itself?

We know the atomic masses of Carbon ($12$) and Oxygen ($16$). The carbonate ion ($C{O}_3^{2-}$) has a mass of $12+(3\times 16)=60{\text{ g mol}}^{-1}$.

If we subtract this from the total molar mass:

While no common alkali metal has an exact atomic mass of $12$, Magnesium ($Mg$) has an atomic mass of $24.3$. However, Magnesium forms $MgC{O}_3$, not $M_{2}C{O}_3$. This tells us that the problem might be purely theoretical or constructed with hypothetical values to test your stoichiometric skills rather than representing a real-world alkali metal carbonate. Always trust the math!