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Ch.4 - Chemical Quantities & Aqueous Reactions
Chapter 4, Problem 68

What is the molarity of ZnCl2 that forms when 25.0 g of zinc completely reacts with CuCl2 according to the following reaction? Assume a final volume of 275 mL. Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)

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Determine the molar mass of zinc (Zn) using the periodic table.
Calculate the moles of zinc (Zn) using the formula: \( \text{moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} \).
Use the stoichiometry of the balanced chemical equation to find the moles of ZnCl_2 produced. Note that the reaction is a 1:1 ratio between Zn and ZnCl_2.
Calculate the molarity of ZnCl_2 using the formula: \( \text{Molarity} = \frac{\text{moles of ZnCl}_2}{\text{volume in liters}} \).
Convert the final volume from milliliters to liters by dividing by 1000.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in units of moles per liter (mol/L). To calculate molarity, one must know the amount of solute in grams, convert it to moles using its molar mass, and then divide by the volume of the solution in liters.
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Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced chemical equations. It allows us to determine the amount of product formed or reactant consumed by using mole ratios derived from the coefficients in the balanced equation. In this case, it helps to relate the amount of zinc used to the amount of ZnCl2 produced.
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Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule. For this problem, knowing the molar mass of zinc (Zn) and zinc chloride (ZnCl2) is essential for converting grams of zinc to moles and subsequently determining the molarity of the resulting solution.
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Open Question
A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq). The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant and the theoretical yield.
Textbook Question

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq) The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the the percent yield.

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Textbook Question

A 55.0-mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs: K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, theoretical yield, percent yield.

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