Question

Calculate the amount of ${\mathrm{CaCO}}_3$ (in grams) that would be required to react with $50.6\mathrm{g}$ of ${\mathrm{SO}}_2$ emitted by a power plant.

Short answer

79.07 grams of CaCO3 are required.

Step-by-step solution

Write the Balanced Chemical Equation

First, we need to consider the chemical reaction between calcium carbonate $({\text{CaCO}}_3)$ and sulfur dioxide $({\text{SO}}_2)$. The balanced chemical equation is:$${\text{CaCO}}_3+{\text{SO}}_2\to {\text{CaSO}}_3+{\text{CO}}_2$$ This equation shows that one mole of calcium carbonate reacts with one mole of sulfur dioxide.

Calculate the Molar Masses

Next, we need the molar masses of ${\text{CaCO}}_3$ and ${\text{SO}}_2$. The molar mass of ${\text{CaCO}}_3$ is calculated as: $${\text{Molar mass of CaCO}}_3=40.08\text{ g/mol (Ca)}+12.01\text{ g/mol (C)}+3\times 16.00\text{ g/mol (O)}=100.09\text{ g/mol}$$ The molar mass of ${\text{SO}}_2$ is calculated as: $${\text{Molar mass of SO}}_2=32.07\text{ g/mol (S)}+2\times 16.00\text{ g/mol (O)}=64.07\text{ g/mol}$$

Convert Grams of SO2 to Moles

We will convert the given mass of ${\text{SO}}_2$ to moles using its molar mass. $${\text{Moles of SO}}_2=\frac{50.6\text{ g}}{64.07\text{ g/mol}}=0.790\text{ mol}$$

Use Stoichiometry to Find Moles of CaCO3 Needed

From the balanced reaction, we know that one mole of ${\text{CaCO}}_3$ reacts with one mole of ${\text{SO}}_2$. Therefore, 0.790 moles of ${\text{SO}}_2$ requires 0.790 moles of ${\text{CaCO}}_3$.

Convert Moles of CaCO3 to Grams

Finally, we convert the moles of ${\text{CaCO}}_3$ to grams using its molar mass.$${\text{Mass of CaCO}}_3=0.790\text{ mol}\times 100.09\text{ g/mol}=79.07\text{ g}$$

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential in stoichiometry. It shows the quantities of reactants and products involved in a reaction. In our example, the reaction between calcium carbonate $({\text{CaCO}}_3)$and sulfur dioxide $({\text{SO}}_2)$gives calcium sulfite$({\text{CaSO}}_3)$and carbon dioxide $({\text{CO}}_2)$.The balanced equation is:$${\text{CaCO}}_3+{\text{SO}}_2\to {\text{CaSO}}_3+{\text{CO}}_2$$
This equation indicates that one mole of ${\text{CaCO}}_3$ reacts with one mole of ${\text{SO}}_2$. The coefficients in a balanced equation provide the ratio necessary to perform stoichiometric calculations. Not balancing the equation means we lack the foundation needed to accurately predict the amounts of reactants and products.

Molar Mass

The molar mass of a compound, necessary for stoichiometric calculations, is the mass of one mole of that compound, expressed in grams per mole (g/mol). To find it, we add together the atomic masses of all the atoms in a molecule. Let's calculate it for ${\text{CaCO}}_3$ and ${\text{SO}}_2$:

• For ${\text{CaCO}}_3$:
• Calcium (Ca) = 40.08 g/mol
• Carbon (C) = 12.01 g/mol
• Oxygen (O has three) = 3 × 16.00 g/mol = 48.00 g/mol
• Total = 100.09 g/mol
• For ${\text{SO}}_2$:
• Sulfur (S) = 32.07 g/mol
• Oxygen (O has two) = 2 × 16.00 g/mol = 32.00 g/mol
• Total = 64.07 g/mol

Understanding molar mass helps convert between grams and moles which is crucial in stoichiometry.

Mole Conversion

Converting between grams and moles involves using the molar mass as a conversion factor. This is a common step in stoichiometric calculations. For instance, to convert 50.6 grams of ${\text{SO}}_2$ to moles, we divide by its molar mass.
The formula is:$$\text{Moles of }{\text{SO}}_2=\frac{\text{Mass of }{\text{SO}}_2(\text{g})}{\text{Molar mass of }{\text{SO}}_2(\text{g/mol})}$$Plugging in the values:$$\text{Moles of }{\text{SO}}_2=\frac{50.6\text{ g}}{64.07\text{ g/mol}}=0.790\text{ mol}$$
Mole conversions are fundamental as they allow us to interrelate different substances in a balanced chemical equation accurately.

Stoichiometric Calculations

Stoichiometric calculations bridge the gap between the balanced chemical equation and the quantitative data from our problem. These calculations ensure you know exactly how much of each reactant is needed or how much product is formed. In our exercise, once you know the moles of ${\text{SO}}_2$, the stoichiometric factor from the balanced equation informs us that 0.790 moles of ${\text{SO}}_2$ require an equal amount of ${\text{CaCO}}_3$.
The next step involves converting moles of ${\text{CaCO}}_3$ back to grams using its molar mass:$${\text{Mass of CaCO}}_3=0.790\text{ mol}\times 100.09\text{ g/mol}=79.07\text{ g}$$
These stoichiometric calculations are critical in real-world applications where accurate measurements are needed, like chemical manufacturing or environmental testing.