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Chapter 4: Problem 70

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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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: \(0.275 \mathrm{L} \times 6.00 \mathrm{M} = 1.65 \mathrm{moles of HCl}\).
03

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 \(0.699 \mathrm{moles of Cl_2} \times \frac{70.90 \mathrm{g}}{1 \mathrm{mol Cl_2} = 49.57 \mathrm{g of Cl_2}\).
04

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 \(0.95 \mathrm{moles of HCl} \times \frac{36.46 \mathrm{g}}{1 \mathrm{mol HCl} = 34.6 \mathrm{g of HCl}\).

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Most popular questions from this chapter

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?

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