Atoms and Molecules - Law of Conservation of Mass

When $100\text{ g}$ of calcium carbonate ($CaCO_3$) is heated, it decomposes to form $56\text{ g}$ of calcium oxide ($CaO$) and some amount of carbon dioxide ($CO_2$). Calculate the mass of carbon dioxide produced using the Law of Conservation of Mass.

According to the Law of Conservation of Mass: $\text{Mass of } CaCO_3 = \text{Mass of } CaO + \text{Mass of } CO_2$ Substituting the given values: $100 = 56 + \text{Mass of } CO_2$ To find the mass of $CO_2$, we subtract the mass of $CaO$ from the total reactant mass: $\begin{array}{r} 100 \\ - 56 \\ \hline 44 \end{array}$ Therefore, the mass of carbon dioxide produced is $44\text{ g}$.

The total mass before the reaction ($100\text{ g}$) must equal the total mass after the reaction. Since we know the mass of one product ($56\text{ g}$), the remaining mass must belong to the other product ($44\text{ g}$).

In a reaction, $5.3\text{ g}$ of sodium carbonate reacts with $6.0\text{ g}$ of ethanoic acid. The products are $2.2\text{ g}$ of carbon dioxide, $0.9\text{ g}$ of water, and $8.2\text{ g}$ of sodium ethanoate. Show that these observations are in agreement with the Law of Conservation of Mass.

First, calculate the total mass of the reactants: $5.3\text{ g} + 6.0\text{ g} = 11.3\text{ g}$. Next, calculate the total mass of the products: $8.2\text{ g} + 2.2\text{ g} + 0.9\text{ g} = 11.3\text{ g}$. Since the total mass of reactants equals the total mass of products ($11.3\text{ g} = 11.3\text{ g}$), the law is verified.

To verify the law, we sum the masses of all chemicals on the left side of the equation and compare it to the sum of the masses of all chemicals on the right side.