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Chapter 4: Problem 107

A \(3.455-\mathrm{g}\) sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If \(0.2815 \mathrm{~g}\) of barium sulfate was obtained, what was the mass percentage of barium in the sample?

Short Answer

Expert verified
The mass percentage of barium in the sample is approximately 4.8%. This is calculated by first finding the moles of barium sulfate obtained and then the mass of barium in the sample. Finally, the mass percentage is determined using the given mass of the mixture.

Step by step solution

01

Identify the molar mass of Barium (Ba) and Barium Sulfate (BaSO4)

To calculate the mass of barium, we need to know the molar mass of barium and barium sulfate. Using the periodic table, we can find the molar mass of each element. Barium (Ba): 137.33 g/mol Sulfur (S): 32.07 g/mol Oxygen (O): 16.00 g/mol Now, let's calculate the molar mass of Barium Sulfate (BaSO4): BaSO4 = 137.33 + (32.07) + 4(16.00) = 137.33 + 32.07 + 64.00 = 233.40 g/mol
02

Find the moles of Barium Sulfate (BaSO4)

Next, we need to find the moles of barium sulfate obtained, using the mass of precipitate and its molar mass. moles of BaSO4 = mass of BaSO4 / molar mass of BaSO4 moles of BaSO4 = 0.2815 g / 233.40 g/mol ≈ 0.00121 mol
03

Find the mass of Barium (Ba) in the sample

Since there is one mole of barium per mole of barium sulfate, the moles of barium will be equal to the moles of barium sulfate. moles of Ba = 0.00121 mol Now we can find the mass of barium from its moles using its molar mass. mass of Ba = moles of Ba × molar mass of Ba mass of Ba = 0.00121 mol × 137.33 g/mol ≈ 0.166 g
04

Calculate the mass percentage of Barium (Ba) in the sample

Now that we have the mass of barium in the sample, we can calculate its mass percentage using the given mass of the mixture. mass percentage of Ba = (mass of Ba / mass of the mixture) × 100 mass percentage of Ba = (0.166 g / 3.455 g) × 100 ≈ 4.8% The mass percentage of barium in the sample is approximately 4.8%.

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Most popular questions from this chapter

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction.

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{3}^{-}\) ?

Explain the following observations: (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\) ions, and yet its aqueous solutions are basic; (b) HF is called a weak acid, and yet it is very reactive; (c) although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) ions than \(\mathrm{SO}_{4}^{2-}\) ions.

A solution is made by mixing \(15.0 \mathrm{~g}\) of \(\mathrm{Sr}(\mathrm{OH})_{2}\) and \(55.0 \mathrm{~mL}\) of \(0.200 \mathrm{MHNO}_{3}\). (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resultant solution acidic or basic?

Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reac- tion. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{AgNO}_{3},\) (b) \(\mathrm{NaNO}_{3}\) and \(\mathrm{NiSO}_{4}\), (c) \(\mathrm{FeSO}_{4}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).
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