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Chapter 4: Problem 72

A sample of \(0.6760 \mathrm{~g}\) of an unknown compound containing barium ions \(\left(\mathrm{Ba}^{2+}\right)\) is dissolved in water and treated with an excess of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). If the mass of the \(\mathrm{BaSO}_{4}\) precipitate formed is \(0.4105 \mathrm{~g}\), what is the percent by mass of Ba in the original unknown compound?

Short Answer

Expert verified
The percent by mass of Barium in the original unknown compound is 35.7%.

Step by step solution

01

Calculation of Molar Masses of Barium and Barium Sulfate

Look up the atomic masses of Barium (Ba), Sulfur (S), and Oxygen (O) in the periodic table. The atomic mass of Barium is 137.327 g/mol, the atomic mass of Sulfur is 32.06 g/mol, and the atomic mass of Oxygen is 16.00 g/mol. To find the molar mass of BaSO4, add the atomic mass of barium to the atomic mass of sulfur and four times the atomic mass of oxygen: \(137.327 \mathrm{~g/mol} + 32.06 \mathrm{~g/mol} + 4(16.00 \mathrm{~g/mol}) = 233.387 \mathrm{~g/mol}\).
02

Calculation of Moles of Ba and BaSO4

We need to find the moles of BaSO4 formed, using the given mass of BaSO4 and its molar mass, the formula is moles = mass / molar mass. So moles of BaSO4 = \(0.4105 \mathrm{~g} / 233.387 \mathrm{~g/mol} = 0.00176 \mathrm{~mol}\). Since the number of moles of Barium in the precipitate will be the same as the moles of BaSO4, moles of Barium in the precipitate is also 0.00176 mol.
03

Calculation of Mass of Barium

The mass of Barium in the sample can be calculated by multiplying the moles of Barium by its atomic mass: \(0.00176 \mathrm{~mol} \times 137.327 \mathrm{~g/mol} = 0.2416 \mathrm{~g}\)
04

Calculate the Percentage of Barium in the Original Compound

Percent by mass = (mass of part / mass of whole) * 100%, so Percent of Ba in the unknown compound = \((0.2416 \mathrm{~g} / 0.6760 \mathrm{~g}) \times 100% = 35.7%\)

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Most popular questions from this chapter

Is it possible to have a reaction in which oxidation occurs and reduction does not? Explain.

Give a chemical explanation for each of these: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. Explain. (b) In the activity series aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. Why? (c) Sodium and potassium lie above copper in the activity series. Explain why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series?

For the complete redox reactions given here, write the half-reactions and identify the oxidizing and reducing agents: (a) \(4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{Cl}_{2}+2 \mathrm{NaBr} \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_{2}\) (c) \(\mathrm{Si}+2 \mathrm{~F}_{2} \longrightarrow \mathrm{SiF}_{4}\) (d) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{HCl}\)

Describe the basic steps involved in an acid-base titration. Why is this technique of great practical value?

How many grams of \(\mathrm{NaCl}\) are required to precipitate most of the \(\mathrm{Ag}^{+}\) ions from \(2.50 \times 10^{2} \mathrm{~mL}\) of \(0.0113 \mathrm{MAgNO}_{3}\) solution? Write the net ionic equation for the reaction.
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