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Chapter 4: Problem 110

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

Short Answer

Expert verified
After the reaction is completed, the molarity of the HCl in the solution is 7.032 M.

Step by step solution

01

Calculation of Initial Moles of HCl

First, find the total mass of the solution, which is \(\text{Volume} \times \(\text{Density} = 1.25 \, \text{L} \times 1.13 \, \text{g/mL} = 1375 \, \text{g}\). Then, determine the mass of HCl in the solution using the mass percent given, which is \(0.257 \times 1375 \, \text{g} = 353.625 \, \text{g HCl}\). The molar mass of HCl is approximately 36.5 g/mol. Thus, initial moles of HCl can be calculated as \( \frac{353.625 \, \text{g}}{36.5 \, \text{g/mol}} = 9.69 \, \text{moles}\).
02

Calculation of Moles of CaCO3

Next, figure out the moles of CaCO3, using its given mass and molar mass, which is approximately 100 g/mol. So, the number moles of CaCO3 = \( \frac{45 \, \text{g}}{100 \, \text{g/mol}} = 0.45 \, \text{moles}\).
03

Determination of Limiting Reactant

From the balanced chemical equation \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{H}_2\text{O} + \text{CaCl}_2 + \text{CO}_2\), it can be seen that one mole of \(\text{CaCO}_3\) reacts with two moles of HCl. Therefore, in order to react with 0.45 moles of \(\text{CaCO}_3\), we would need \(0.45 \times 2 = 0.9\) moles of HCl. But 9.69 moles are available, so HCl is in excess, and \(\text{CaCO}_3\) is limiting. Determination of the limiting reactant is crucial as it limits the progression of the reaction.
04

Calculation of Final moles of HCl

As \(\text{CaCO}_3\) is the limiting reactant, all of it will react with HCl. Knowing the stoichiometry, 0.9 moles of HCl will react with \(\text{CaCO}_3\). Thus, subtract the moles of HCl reacted from the initial moles. Therefore, final moles of HCl = Initial moles - reacted moles = 9.69 - 0.9 = 8.79 moles.
05

Calculation of Final Molarity of HCl

Since the volume of the solution remains constant (1.25 L), the final molarity of HCl = moles of HCl / volume of the solution in liters. Therefore, Molarity = \(\frac{8.79 \, \text{moles}}{1.25 \, \text{L}} = 7.032 \, \text{M}\).

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Most popular questions from this chapter

To obtain a solution that is \(1.00 \mathrm{M} \mathrm{NaNO}_{3}\), you should prepare (a) 1.00 L of aqueous solution containing \(100 \mathrm{g} \mathrm{NaNO}_{3} ;\) (b) \(1 \mathrm{kg}\) of aqueous solution containing \(85.0 \mathrm{g} \mathrm{NaNO}_{3} ;(\mathrm{c}) 5.00 \mathrm{L}\) of aqueous solution containing \(425 \mathrm{g} \mathrm{NaNO}_{3} ;(\mathrm{d})\) an aqueous solution containing \(8.5 \mathrm{mg} \mathrm{NaNO}_{3} / \mathrm{mL}\).

Under appropriate conditions, copper sulfate, potassium chromate, and water react to form a product containing \(\mathrm{Cu}^{2+},\) \(\mathrm{CrO}_{4}{^2}{^-},\) and \(\mathrm{OH}^{-}\) ions. Analysis of the compound yields \(48.7 \% \mathrm{Cu}^{2+}, 35.6 \% \mathrm{CrO}_{4}{^2}{-},\) and \(15.7 \% \mathrm{OH}^{-}\). (a) Determine the empirical formula of the compound. (b) Write a plausible equation for the reaction.

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