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Consider the reaction below. \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) How many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react completely with \(415 \mathrm{mL}\) of \(0.477 \mathrm{M} \mathrm{HCl} ?\) (b) How many kilograms of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react with 324 L of a HCl solution that is 24.28\% HCl by mass, and has a density of \(1.12 \mathrm{g} / \mathrm{mL} ?\)

Short Answer

Expert verified
a) 7.33 grams of Ca(OH)2 are needed to react completely with 415 ml of 0.477 M HCl. b) 8.90 kilograms of Ca(OH)2 are required to react with 324 L of a HCl solution that is 24.28% HCl by mass, and has a density of 1.12 g/mL.

Step by step solution

01

Determine the reactant moles from the given volume and molarity

From the molarity and volume of the HCl solution, calculate the moles. Molarity = moles/volume. So, moles of HCl = Molarity * Volume = 0.477 M * 0.415 L = 0.19785 moles.
02

Use stoichiometry to find the moles of Ca(OH)2

From the reaction, 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Hence moles of Ca(OH)2 = 0.19785 moles / 2 = 0.098925 moles.
03

Convert moles to grams

Convert the moles of Ca(OH)2 to grams using its molar mass (Molar mass of Ca(OH)2 = 74.093 g/mol). Grams = moles * Molar mass = 0.098925 moles * 74.093 g/mol = 7.33 g.
04

Determine the mass of HCl from the given solution volume and density

The solution volume is given, and the percentage composition by mass and density of the solution is known. The mass of the HCl in the solution is thus = Volume of solution * Density * (percentage composition / 100) = 324 L * (1.12 g/mL) * (24.28 / 100) = 8766.7712 g or 8.77 kg.
05

Convert mass to moles

Convert the mass of HCl to moles using its molar mass (Molar mass of HCl = 36.461 g/mol). Moles of HCl = Mass / Molar Mass = 8766.7712 g / 36.461 g/mol = 240.409 moles.
06

Use stoichiometry to find the moles of Ca(OH)2

Using the stoichiometric coefficients from the balanced chemical equation, find the moles of Ca(OH)2. Moles of Ca(OH)2 = Moles of HCl / 2 = 240.409 moles / 2 = 120.205 moles.
07

Convert moles to kilograms

Convert the moles of Ca(OH)2 to kilograms using its molar mass. Kilograms of Ca(OH)2 = Moles * Molar mass / 1000 = 120.205 moles * 74.093 g/mol / 1000 = 8.90 kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
In a chemical reaction, substances known as reactants undergo a transformation to form new substances called products. This occurs through the breaking and formation of chemical bonds. Consider the reaction: \( \text{Ca(OH)}_2(s) + 2 \text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + 2 \text{H}_2\text{O}(l) \). Here, calcium hydroxide \( \text{Ca(OH)}_2 \) reacts with hydrochloric acid \( \text{HCl} \) to produce calcium chloride \( \text{CaCl}_2 \) and water \( \text{H}_2\text{O} \). Even though these changes seem simple, a deeper understanding requires us to examine the stoichiometry of the reaction. Understanding the stoichiometric coefficients, which is the number before each substance in the balanced chemical equation, is crucial. These coefficients indicate the ratio of moles that react or are produced. For this specific reaction:
  • 1 mole of \( \text{Ca(OH)}_2 \) reacts with 2 moles of \( \text{HCl} \).
  • This means for every mole of calcium hydroxide, it requires two moles of hydrochloric acid.
A balanced equation ensures that the law of conservation of mass is satisfied; the number of each type of atom should be the same on both sides of the reaction.
Molarity
Molarity is a measure of concentration used in chemistry. It is defined as the number of moles of solute (the substance being dissolved) in one liter of solution. The formula for calculating molarity is:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] In the given reaction, molarity helps us calculate the moles of \( \text{HCl} \) in the solution. For example, if you have 0.477 M \( \text{HCl} \) and a volume of 0.415 L, you can find the moles as follows:- Moles of \( \text{HCl} \) = 0.477 M * 0.415 L = 0.19785 moles. This conversion from volume to moles is vital because chemical reactions occur based on moles, not on a simple volume or mass. Using molarity is essential in converting measurable volumes into usable data for stoichiometric calculations. By knowing the concentration and volume, chemists can predict how much product will form or how much of a reactant is needed for the reaction.
Chemical Calculations
Chemical calculations, often involving stoichiometry, allow scientists to predict the outcomes of reactions. They help determine the amounts of reactants required or products formed in a reaction. The step-by-step calculations provided can guide us in solving such chemistry problems. Let's take problem (b) as an illustration:
  • We first find the mass of \( \text{HCl} \) in 324 L of solution using density and percentage composition: \( 324 \text{ L} \times 1.12 \text{ g/mL} \times \frac{24.28}{100} = 8766.77 \text{ g} \).
  • Next, using molar mass, we find moles: \( \frac{8766.77 \text{ g}}{36.461 \text{ g/mol}} = 240.409 \text{ moles} \).
  • Then, stoichiometry helps us find moles of \( \text{Ca(OH)}_2 \): \( \frac{240.409}{2} = 120.205 \text{ moles} \).
  • Finally, we convert moles of \( \text{Ca(OH)}_2 \) to kilograms: \( 120.205 \times 74.093 \text{ g/mol} / 1000 = 8.90 \text{ kg} \).
These calculations are key in converting theoretical equations into practical quantities, ensuring the precise measurement needed for chemical synthesis and experiments.

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Most popular questions from this chapter

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.

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