Question

Calcium carbonate, \(\mathrm{CaCO}_{2}\), is often used in commercial antacids. It acts to reduce the acidity in the stomach by neutralizing stomach acid, which is mostly HCL, by the following reaction: $$ \mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(d) $$ What mass of \(\mathrm{CaCO}_{3}\) is needed to neutralize \(0.020 \mathrm{~mol} \mathrm{HCl}\) ?

Short answer

The mass of calcium carbonate (\(\mathrm{CaCO_3}\)) needed to neutralize 0.020 mol of hydrogen chloride (\(\mathrm{HCl}\)) is approximately 1.001 g.

Step-by-step solution

Analyze the Balanced Chemical Reaction

The balanced chemical reaction is: \(\mathrm{CaCO}_{3}\)(s)+2\(\mathrm{HCl}\)(aq) \(\longrightarrow\) \(\mathrm{CaCl}_{2}\)(aq) + \(\mathrm{CO}_{2}\)(g) + \(\mathrm{H}_{2}\) \(\mathrm{O}\)(l). From this reaction, we can see that one mole of calcium carbonate neutralizes two moles of hydrochloric acid. This will be useful for calculating the amount of calcium carbonate needed.

Calculate the Amount of Calcium Carbonate Needed

Since 1 mol of calcium carbonate neutralizes 2 mol of HCl, the moles of calcium carbonate needed to neutralize 0.020 mol of HCl are \(\frac{0.020\text{ mol HCl}}{2} = 0.010\text{ mol CaCO}_3\). Now, to find the mass of calcium carbonate needed, we multiply the number of moles by the molar mass of calcium carbonate, which is approximately 100.1 g/mol, \(0.010\text{ mol CaCO}_3 \times 100.1\text{ g/mol} = 1.001\text{ g}\).

Final Calculation

The final result to the problem; it was found that the mass of calcium carbonate needed to neutralize 0.020 mol of HCl is approximately 1.001 g.

Key concepts

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the amounts of reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, the mass and the number of atoms of each element must be the same on both sides of a balanced equation.

For instance, in our exercise, we are dealing with the neutralization of stomach acid (HCl) using calcium carbonate (CaCO₃). The balanced equation shows that each molecule of CaCO₃ neutralizes two molecules of HCl. Utilizing stoichiometry, we can determine the exact amount of CaCO₃ needed to neutralize a specific amount of HCl. This calculation involves converting moles to grams, which hinges on the concept known as the molar mass.

Chemical Reaction Balancing

Chemical reaction balancing is a critical step in stoichiometry, as it ensures that the number of atoms for each element is equal on both sides of the reaction. This is necessary because of the law of conservation of mass. To balance a reaction, one must adjust the coefficients (the numbers in front of compounds) until the number of atoms of each element is the same on both the reactant and product sides of the equation.

In our example, the balanced reaction is \(\mathrm{CaCO}_{3}(s) + 2\mathrm{HCl}(aq) \longrightarrow \mathrm{CaCl}_{2}(aq) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\), showing that one mole of CaCO₃ reacts with two moles of HCl to produce one mole of CaCl₂, one mole of CO₂, and one mole of H₂O. This indicates a 1:2 molar ratio between CaCO₃ and HCl, which is key to performing stoichiometric calculations.

Molar Mass Calculation

The molar mass of a substance is the mass in grams of one mole of that substance. It is a physical property that is used to convert moles of a substance to grams, which is an essential step in stoichiometry for many chemistry problems, including the neutralization reaction in our exercise.

The molar mass is calculated by summing the atomic masses of all the atoms in a molecule. For example, calcium carbonate (CaCO₃) has a molar mass of approximately 100.1 g/mol. This value comes from adding the atomic mass of one calcium atom (about 40.1 g/mol), one carbon atom (about 12.0 g/mol), and three oxygen atoms (about 16.0 g/mol each).

In the final calculation step from our exercise, we use the molar mass of CaCO₃ to convert moles of CaCO₃ to grams. We obtain the mass required to neutralize 0.020 mol of HCl by using the molar ratio derived from the balanced equation and the molar mass of CaCO₃. We found that you need approximately 1.001 grams of CaCO₃ to neutralize 0.020 mol of HCl.