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Chapter 4: Problem 109

A 99.8 mL sample of a solution that is \(120 \%\) KI by mass \((d=1.093 \mathrm{g} / \mathrm{mL})\) is added to \(96.7 \mathrm{mL}\) of another solution that is \(14.0 \% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass \((d=1.134 \mathrm{g} / \mathrm{mL})\) How many grams of \(\mathrm{PbI}_{2}\) should form? \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{KI}(\mathrm{aq}) \longrightarrow \mathrm{PbI}_{2}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq})\)

Short Answer

Expert verified
The reaction should form approximately 21.19 g of PbI2

Step by step solution

01

Conversion of Solution Volumes to Masses

Use the provided densities and volumes to calculate the mass of each solution.\n Mass of KI solution = \(\(99.8 mL\) * \(1.093 g/mL\) = \(109.15 g\)\).\n Mass of \(Pb(NO_3)_2\) solution = \(\(96.7 mL\) * \(1.134 g/mL\) = \(109.67 g\)\).
02

Determination of Mass of Reactants

Utilize the mass percentages to find the mass of each reactant.\n Mass of KI = \(\(120 \%\) * \(109.15 g\) / \(100 \%\) = \(130.98 g\)\).\n Mass of \(Pb(NO_3)_2\) = \(\(14.0 \%\) * \(109.67 g\) / \(100 \%\) = \(15.35 g\)\).
03

Conversion of Mass to Moles

Use the molecular masses to convert the mass of the reactants to moles. For KI, molecular mass is \(166.0 g/mol\), and for \(Pb(NO_3)_2\), molecular mass is \(331.2 g/mol\).\nMoles of KI = \(\(130.98 g\) / \(166.0 g/mol\) = \(0.789 mol\)\) \nMoles of \(Pb(NO_3)_2\) = \(\(15.35 g\) / \(331.2 g/mol\) = \(0.046 mol\) \), Note that the reaction consumes KI and \(Pb(NO_3)_2\) in a 2:1 ratio.
04

Identify Limiting Reactant

The limiting reactant is the reactant that is completely used up in the reaction first. With a 2:1 consumption ratio and given the amount of both KI and \(Pb(NO_3)_2\), it is obvious that \(Pb(NO_3)_2\) is the limiting reactant because its quantity is less than half of the quantity of KI.
05

Calculation of Mass of PbI2 formed

The equation shows that 1 mole of \(Pb(NO_3)_2\) produces 1 mole of PbI2. Hence moles of PbI2 produced = moles of \(Pb(NO_3)_2\), which equals \(0.046 mol\). Using the molecular mass of PbI2, \(460.6 g/mol\), calculate the mass of PbI2 produced. \n Mass of PbI2 = Moles of PbI2 * molar mass of PbI2 = \(0.046 mol\) * \(460.6 g/mol\) = \(21.19 g\) of PbI2

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Most popular questions from this chapter

Given two liters of \(0.496 \mathrm{M} \mathrm{KCl},\) describe how you would use this solution to prepare \(250.0 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{KCl} .\) Give sufficient details so that another student could follow your instructions.

A laboratory method of preparing \(\mathrm{O}_{2}(\mathrm{g})\) involves the decomposition of \(\mathrm{KClO}_{3}(\mathrm{s})\) $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) can be produced by the decomposition of \(32.8 \mathrm{g} \mathrm{KClO}_{3} ?\) (b) How many grams of \(\mathrm{KClO}_{3}\) must decompose to produce \(50.0 \mathrm{g} \mathrm{O}_{2} ?\) (c) How many grams of KCl are formed, together with \(28.3 \mathrm{g} \mathrm{O}_{2},\) in the decomposition of \(\mathrm{KClO}_{3} ?\)

How many milliliters of \(0.650 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) are needed to precipitate all the silver in \(415 \mathrm{mL}\) of \(0.186 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) as \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s}) ?\) \(2 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq}) $$

Briefly describe (a) balancing a chemical equation; (b) preparing a solution by dilution; (c) determining the limiting reactant in a reaction.

The reaction of calcium hydride and water produces calcium hydroxide and hydrogen as products. How many moles of \(\mathrm{H}_{2}(\mathrm{g})\) will be formed in the reaction between \(0.82 \mathrm{mol} \mathrm{CaH}_{2}(\mathrm{s})\) and \(1.54 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ?\)
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