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Chapter 4: Problem 68

Lithopone is a brilliant white pigment used in waterbased interior paints. It is a mixture of \(\mathrm{BaSO}_{4}\) and \(\mathrm{ZnS}\) produced by the reaction $$\mathrm{BaS}(\mathrm{aq})+\mathrm{ZnSO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{ZnS}(\mathrm{s})+\mathrm{BaSO}_{4}(\mathrm{s})$$ How many grams of lithopone are produced in the reaction of \(315 \mathrm{mL}\) of \(0.275 \mathrm{M} \mathrm{ZnSO}_{4}\) and \(285 \mathrm{mL}\) of 0.315 M BaS?

Short Answer

Expert verified
The total mass of Lithopone produced in the reaction is 28.66 grams.

Step by step solution

01

Determine the moles of reactants

First we need to determine the amount of the reactants in moles. For this we multiply the volume of each reactant in liters with their corresponding molarity given as \(M\). Therefore, For \(ZnSO_{4}\): \(0.315L\) * \(0.275M\) = \(0.086625 mol\)For \(BaS\): \(285mL = 0.285L\), \(0.285L\) * \(0.315M\) = \(0.089775 mol\)
02

Identify the Limiting Reactant

Since the reaction uses one mole of each reactant to produce one mole of each product, the limiting reactant is \(ZnSO_{4}\) because it has fewer moles available, \(0.086625 mol\), compared to \(BaS\) which has \(0.089775 mol\). Hence, only \(0.086625 mol\) each of \(ZnS\) and \(BaSO_{4}\) can be formed.
03

Convert Moles of product to Grams

Finally, convert moles of chemical products to grams using their molar masses. For \(ZnS\): \(0.086625 mol\) * \(97.47 g/mol = 8.44g\)For \(BaSO_{4}\): \(0.086625 mol\) * \(233.4 g/mol = 20.22g\)
04

Calculate Total Mass of Lithopone

As Lithopone is the mixture of the two products, we would have to add up the masses of \(ZnS\) and \(BaSO_{4}\) which gives us \(8.44g + 20.22g = 28.66g\)

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Most popular questions from this chapter

What are the molarities of the following solutes when dissolved in water? (a) \(2.92 \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH}\) in 7.16 L of solution (b) \(7.69 \mathrm{mmol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) in \(50.00 \mathrm{mL}\) of solution (c) \(25.2 \mathrm{g} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) in \(275 \mathrm{mL}\) of solution

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 \(\mathrm{KClO}_{3} ?\)

Baking soda, \(\mathrm{NaHCO}_{3}\), is made from soda ash, a common name for sodium carbonate. The soda ash is obtained in two ways. It can be manufactured in a process in which carbon dioxide, ammonia, sodium chloride, and water are the starting materials. Alternatively, it is mined as a mineral called trona (left photo). Whether the soda ash is mined or manufactured, it is dissolved in water and carbon dioxide is bubbled through the solution. Sodium bicarbonate precipitates from the solution. As a chemical analyst you are presented with two samples of sodium bicarbonate-one from the manufacturing process and the other derived from trona. You are asked to determine which is purer and are told that the impurity is sodium carbonate. You decide to treat the samples with just sufficient hydrochloric acid to convert all the sodium carbonate and bicarbonate to sodium chloride, carbon dioxide, and water. You then precipitate silver chloride in the reaction of sodium chloride with silver nitrate. A \(6.93 \mathrm{g}\) sample of baking soda derived from trona gave \(11.89 \mathrm{g}\) of silver chloride. A \(6.78 \mathrm{g}\) sample from manufactured sodium carbonate gave \(11.77 \mathrm{g}\) of silver chloride. Which sample is purer, that is, which has the greater mass percent \(\mathrm{NaHCO}_{3} ?\)

Given a \(0.250 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution, describe how you would prepare a solution that is \(0.0125 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{CrO}_{4} .\) That is, what combination(s) of pipet and volumetric flask would you use? Typical sizes of volumetric flasks found in a general chemistry laboratory are \(100.0,250.0,500.0,\) and \(1000.0 \mathrm{mL},\) and typical sizes of volumetric pipets are 1.00,5.00,10.00 \(25.00,\) and \(50.00 \mathrm{mL}\)

Chromium(II) sulfate, \(\mathrm{CrSO}_{4^{\prime}}\) is a reagent that has been used in certain applications to help reduce carbon-carbon double bonds \((\mathrm{C}=\mathrm{C})\) in molecules to single bonds ( \(\mathrm{C}-\mathrm{C}\) ). The reagent can be prepared via the following reaction. $$\begin{array}{c} 4 \mathrm{Zn}(\mathrm{s})+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{ZnSO}_{4}(\mathrm{aq})+2 \mathrm{CrSO}_{4}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$
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