Problem 70 - General Chemistry: Principles and Modern Applications

Question

Chlorine can be generated by heating together calcium hypochlorite and hydrochloric acid. Calcium chloride and water are also formed. (a) If \(50.0 \mathrm{g}\) \(\mathrm{Ca}(\mathrm{OCl})_{2}\) and \(275 \mathrm{mL}\) of \(6.00 \mathrm{M} \mathrm{HCl}\) are allowed to react, how many grams of chlorine gas will form? (b) Which reactant, \(\mathrm{Ca}(\mathrm{OCl})_{2}\) or \(\mathrm{HCl}\), remains in excess, and in what mass?

Step-by-Step Solution

Answer

49.57 grams of chlorine gas will form. Hydrochloric acid (HCl) will be the excess reactant with a mass of 34.6 grams remaining.

Step 1: Balance the chemical equation

The balanced chemical equation for this reaction is \(2 \mathrm{HCl} + \mathrm{Ca(OCl)_2} \rightarrow 2 \mathrm{Cl_2} + \mathrm{CaCl_2} + 2 \mathrm{H_2O}\). This indicates that for every one mole of calcium hypochlorite, two moles of hydrochloric acid are required, and two moles of chlorine gas are produced.

Step 2: Calculate the number of moles of each reactant

Convert the mass of calcium hypochlorite to moles using its molar mass, (\(1 \mathrm{mol Ca(OCl)_2} = 142.98 \mathrm{g}\), yielding \(50.0 \mathrm{g} \times \frac{1 \mathrm{mol}}{142.98 \mathrm{g}} = 0.349 \mathrm{moles of Ca(OCl)_2}\). Hydrochloric acid’s molarity is given so its moles can be calculated: \(0.275 \mathrm{L} \times 6.00 \mathrm{M} = 1.65 \mathrm{moles of HCl}\).

Step 3: Calculate the expected mass of chlorine gas

According to the balanced equation, one mole of calcium hypochlorite produces two moles of chlorine gas. Therefore, \(0.349 \mathrm{moles of Ca(OCl)_2} \times \frac{2 \mathrm{moles of Cl_2}}{1 \mathrm{mol Ca(OCl)_2}} = 0.699 \mathrm{moles of Cl_2}\). Converting moles of chlorine to grams using its molar mass (\(1 \mathrm{mol Cl_2} = 70.90 \mathrm{g}\) gives \(0.699 \mathrm{moles of Cl_2} \times \frac{70.90 \mathrm{g}}{1 \mathrm{mol Cl_2} = 49.57 \mathrm{g of Cl_2}\).

Step 4: Identify the excess reactant and its mass

Based on the balanced equation, one mole of calcium hypochlorite requires two moles of hydrogen chloride. Hence, the 1.65 moles of HCl are more than enough to react with 0.349 moles of Ca(OCl)_2, so HCl is the excess reactant. The moles of HCl left over can be calculated: \(1.65 \mathrm{moles of HCl} - 0.349 \mathrm{moles of Ca(OCl)_2} \times \frac{2 \mathrm{moles of HCl}}{1 \mathrm{mol Ca(OCl)_2} = 0.95 \mathrm{moles of HCl}\). Converting moles of HCl to grams, using its molar mass (\(1 \mathrm{mol HCl} = 36.46 \mathrm{g}\), results in \(0.95 \mathrm{moles of HCl} \times \frac{36.46 \mathrm{g}}{1 \mathrm{mol HCl} = 34.6 \mathrm{g of HCl}\).

Key Concepts

Balanced Chemical Equation

In chemistry, understanding the stoichiometry begins with a balanced chemical equation. For our given reaction, the balanced equation is: \[ 2 \mathrm{HCl} + \mathrm{Ca(OCl)_2} \rightarrow 2 \mathrm{Cl_2} + \mathrm{CaCl_2} + 2 \mathrm{H_2O} \] This equation tells us the relationship between reactants and products. Each compound's coefficients give the number of moles involved. For every one mole of calcium hypochlorite (\(\mathrm{Ca(OCl)_2}\)), two moles of hydrochloric acid (\(\mathrm{HCl}\)) are required. This also yields two moles of chlorine gas, while forming calcium chloride and water as byproducts.
Understanding this balanced setup is vital, as it lays the groundwork for mole calculations and determining which reactant is in excess. It ensures that all atoms are accounted for and conserved across each side of the equation, abiding by the law of conservation of mass.

Mole Calculations

Mole calculations bridge the macroscopic world we see with the microscopic world of atoms and molecules. It's like translating mass into moles, which are the chemist's way of counting particles.
The molar mass of a compound helps convert the mass of substances into moles. Using given data, we calculate the moles of each reactant. For calcium hypochlorite, \(1\, \mathrm{mol}\, \mathrm{Ca(OCl)_2}\) equals 142.98 grams. Thus, \(50.0\, \mathrm{g} \times \frac{1\, \mathrm{mol}}{142.98\, \mathrm{g}} = 0.349\, \mathrm{mol}\) of \(\mathrm{Ca(OCl)_2}\).
For hydrochloric acid with a molarity of \(6.00\, \mathrm{M}\), the moles are \(0.275\, \mathrm{L} \times 6.00\, \mathrm{M} = 1.65\, \mathrm{mol}\, \mathrm{HCl}\). Mole calculations enable us to predict product formation and help analyze which reactant gets consumed fully or remains in excess.

Excess Reactant

Understanding which reactant is in excess helps optimize chemical reactions, preventing wastage of materials.
In our balanced equation, \(\mathrm{Ca(OCl)_2}\) requires \(2\, \mathrm{moles}\) of \(\mathrm{HCl}\) per mole. With \(0.349\, \mathrm{mol}\, \mathrm{Ca(OCl)_2}\) reacting, theoretically \(0.698\, \mathrm{mol}\, \mathrm{HCl}\) is required. However, since \(1.65\, \mathrm{mol}\, \mathrm{HCl}\) is available, there’s more than enough hydrogen chloride present.
Therefore, \(\mathrm{HCl}\) is the excess reactant. To find out how much \(\mathrm{HCl}\) remains, we subtract the moles that reacted: \(1.65\, \mathrm{mol}\, \mathrm{HCl} - 0.698\, \mathrm{mol} = 0.952\, \mathrm{mol}\) of \(\mathrm{HCl}\). Converting this into mass gives \(0.952\, \mathrm{mol} \times 36.46\, \mathrm{g/mol} = 34.68\, \mathrm{g}\).
This insight is crucial for applications where efficiency and cost-effectiveness are important, such as in industrial processes. By managing excess reactants, industries can minimize waste and maximize output.