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Chapter 4: Problem 106

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} .\) A \(0.456-\mathrm{g}\) sample of the mixture is dissolved in water, and an excess of \(\mathrm{NaOH}\) is added. producing a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{3} .\) The precipitate is filtered. dried, and weighed. Its mass is \(0.107 \mathrm{~g}\). Calculate the following. a. the mass of iron in the sample b. the mass of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample c. the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample.,

Short Answer

Expert verified
a. Mass of Iron in the sample can be calculated as: Moles of Fe(OH)₃ = \(\frac{0.107 g}{106.87 \frac{g}{mol}} = 0.001 \: mol\) Mass of Iron = Moles of Iron × molar mass of Iron = \(0.001 \: mol \times 55.85 \frac{g}{mol} = 0.05585 g\) b. Mass of Fe(NO₃)₃ in the sample: Mass of Fe(NO₃)₃ = Moles of Iron × Molar mass of Fe(NO₃)₃ = \(0.001 \: mol \times 241.85 \frac{g}{mol} = 0.24185 g\) c. Mass percent of Fe(NO₃)₃ in the sample: Mass percent of Fe(NO₃)₃ = \(\frac{0.24185 g}{0.456\:g}\) × 100 = 53.04%

Step by step solution

01

Find the moles of Fe(OH)₃ precipitate

Firstly, we need to find the moles of Fe(OH)₃ precipitate. We can do this by dividing the given mass of Fe(OH)₃ by the molar mass of Fe(OH)₃. The molar mass of Fe(OH)₃ is: \(1\times(55.85 \frac{g}{mol})+3\times(15.999 \frac{g}{mol}+1.007 \frac{g}{mol}) = 106.87 \frac{g}{mol}\). Calculate the moles of Fe(OH)₃ precipitate: Moles of Fe(OH)₃ = \(\frac{0.107 g}{106.87 \frac{g}{mol}}\)
02

Find the moles of Iron in the sample

Since there is one mole of Iron in each mole of Fe(OH)₃, the moles of Fe(OH)₃ precipitate are equal to the moles of Iron in the sample. Moles of Iron = Moles of Fe(OH)₃
03

Calculate the mass of Iron in the sample

Now that we have the moles of Iron, we can find the mass of Iron in the sample by multiplying the moles of Iron by the molar mass of Iron. The molar mass of Iron is 55.85 g/mol. Mass of Iron = Moles of Iron × molar mass of Iron
04

Calculate the mass of Fe(NO₃)₃ in the sample

Knowing that the mass of Iron found is from Fe(NO₃)₃, we can calculate the mass of Fe(NO₃)₃ in the sample by using stoichiometry. Calculate the molar mass of Fe(NO₃)₃: \(1 \times (55.85 \frac{g}{mol}) + 3 \times (14.0067 \frac{g}{mol} + 3 \times (15.999 \frac{g}{mol}) = 241.85 \frac{g}{mol}\). Mass of Fe(NO₃)₃ = Moles of Iron × Molar mass of Fe(NO₃)₃
05

Calculate the mass percent of Fe(NO₃)₃ in the sample

Finally, we can calculate the mass percent of Fe(NO₃)₃ in the sample by dividing the mass of Fe(NO₃)₃ by the total mass of the sample (0.456 g) and then multiplying by 100. Mass percent of Fe(NO₃)₃ = \(\frac{Mass\:of\:Fe(NO_{3})_{3}}{0.456\:g}\) × 100 Following these steps, we can calculate the mass of Iron, the mass of Fe(NO₃)₃, and the mass percent of Fe(NO₃)₃ in the sample.

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Most popular questions from this chapter

Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion d. sulfate ion b. calcium ion e. mercury(I) ion, \(\mathrm{Hg}_{2}^{2+}\) c. iron(III) ion f. silver ion

What volume of each of the following acids will react completely with \(50.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH} ?\) a. \(0.100 \mathrm{M} \mathrm{HCl}\) b. \(0.150 \mathrm{MHNO}_{3}\) c. \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) ( 1 acidic hydrogen)

An average human being has about \(5.0 \mathrm{~L}\) of blood in his or her body. If an average person were to eat \(32.0 \mathrm{~g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{~g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 \mathrm{M} \mathrm{NaOH}\) from \(1.00 \mathrm{M} \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?
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