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Chapter 17: Problem 85

Zinc can be determined volumetrically by the precipitatio reactions, $$ 3 \mathrm{Zn}^{2+}+2 \mathrm{~K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \longrightarrow \mathrm{K}_{2} \mathrm{Zn}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+6 \mathrm{~K}^{+} $$ A sample of Zn ore weighing \(1.5432 \mathrm{~g}\) was prepared for reaction and required \(34.68 \mathrm{~mL}\) of \(0.1043 \mathrm{M} \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) for titration. What is percentage of zinc in the ore?

Short Answer

Expert verified
The percentage of zinc in the ore is approximately 22.97%.

Step by step solution

01

Write the Reaction and Mole Ratios

The given chemical reaction is \( 3 \mathrm{Zn}^{2+} + 2 \mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6] \rightarrow \mathrm{K}_2\mathrm{Zn}_3[\mathrm{Fe}(\mathrm{CN})_6] + 6 \mathrm{K}^+ \). From this reaction, we see that 3 moles of \(\mathrm{Zn}^{2+}\) react with 2 moles of \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\). Thus, the mole ratio of \(\mathrm{Zn}^{2+}\) to \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\) is 3:2.
02

Calculate Moles of Titrant Used

The formula for moles is volume (L) \( \times \) molarity. The titrant is \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\), with a volume of 34.68 mL, which is \(0.03468\) L, and a molarity of 0.1043 M. Thus, the moles of \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\) used are \(0.03468 \times 0.1043 = 0.003617324\) moles.
03

Calculate Moles of Zn

From Step 1, the mole ratio of \(\mathrm{Zn}^{2+}\) to \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\) is 3:2. Therefore, the moles of \(\mathrm{Zn}^{2+}\) is \( \frac{3}{2} \times 0.003617324 = 0.005425986\) moles.
04

Calculate Mass of Zn

Use the moles of \(\mathrm{Zn}^{2+}\) to find its mass. The molar mass of zinc (Zn) is approximately 65.38 g/mol. Thus, the mass of Zn is \(0.005425986 \times 65.38 = 0.354609\) g.
05

Calculate Percentage of Zn in Ore

The percentage of Zn in the ore is given by the formula: \(100 \times \frac{\text{mass of Zn}}{\text{mass of ore}}\). The mass of the ore is 1.5432 g. Thus, the percentage of Zn is \(100 \times \frac{0.354609}{1.5432} \approx 22.97\%\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reactions
Precipitation reactions are chemical reactions where two soluble substances react to form an insoluble solid, known as a precipitate. In the context of zinc, the reaction involves zinc ions and potassium ferrocyanide. When zinc ions combine with potassium ferrocyanide, it forms an insoluble complex, precipitating out of the solution. Some key points to remember about precipitation reactions include:
  • They generally involve reactions between ionic compounds in aqueous solutions.
  • The formation of a solid precipitate is usually accompanied by a color change.
  • Precipitation reactions can be used for analytical purposes, as seen in the determination of zinc content through volumetric analysis.
Understanding precipitation reactions helps in identifying possible outcomes when mixing two solutions and is fundamental in the separation and qualitative analysis of ions.
Mole Ratios
Mole ratios are essential for stoichiometry and are derived from balanced chemical equations. They provide a quantitative relationship between reactants and products. In our zinc precipitation reaction, the equation represents three moles of zinc ions reacting with two moles of potassium ferrocyanide. This gives a mole ratio of 3:2 for zinc to potassium ferrocyanide. Using mole ratios, chemists can:
  • Determine how much of each reactant is required to react completely.
  • Calculate the amount of product produced in a reaction.
  • Examine reaction yields and efficiencies.
Having a grasp on mole ratios allows you to predict the quantities involved in a chemical reaction, highlighting their significance in both laboratory and industrial applications.
Molarity
Molarity, denoted as M, is a way to express concentration. It is defined as the number of moles of solute per liter of solution. In our example, the molarity of potassium ferrocyanide is 0.1043 M. This means there are 0.1043 moles of this compound in every liter of the solution.Calculating molarity involves:
  • Measuring the volume of the solution accurately.
  • Knowing the number of moles of solute present.
  • Using the formula: \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]
Molarity is a crucial concept in volumetric analysis, helping to perform accurate titrations and determine concentrations needed for reactions. It aids in understanding how concentrated a solution is and in adjusting conditions to achieve desired reactions.
Percentage Composition
Percentage composition refers to the percentage by mass of each element in a compound or mixture. In the context of the given problem, it involves calculating how much zinc is present in zinc ore.To determine percentage composition, follow these steps:
  • First, find the mass of the element of interest, in this case, zinc.
  • Next, divide the mass of zinc by the total mass of the ore.
  • Finally, multiply the result by 100 to get a percentage.\[\text{Percentage} = 100 \times \frac{\text{mass of Zn}}{\text{mass of ore}}\]
Understanding percentage composition is important in analyzing the makeup of different substances, especially in determining the purity of metals extracted from ores. It helps in quality control and ensuring the material meets specified standards.

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Most popular questions from this chapter

\(1.6 \mathrm{~g}\) of pyrolusite was treated with \(60 \mathrm{~mL}\) of normal oxalic acid and some \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The oxalic acid left undecomposed was made upto \(250 \mathrm{~mL}, 25 \mathrm{~mL}\) of this solution required \(32 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) potassium permanganate \(\left(\mathrm{KMnO}_{4}\right) .\) Calculate the percentage of pure \(\mathrm{MnO}_{2}\) in pyrolusite.

\(1.13 \mathrm{~g}\) of an ammonium sulphate were treated with \(50 \mathrm{~mL}\) of normal \(\mathrm{NaOH}\) solution and boiled till no more ammonia was given off. The excess of the alkali solution left over was titrated with normal \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The volume required was \(30 \mathrm{~mL}\). Find out the percentage of \(\mathrm{NH}_{3}\) in the salt.

6 g sample of \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}+\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) mixture and an inert material was treated with excess of aqueous \(\mathrm{KI}\) which reduces all \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) ion. The resulting solution was diluted to \(50 \mathrm{~mL}\). \(10 \mathrm{~mL}\) of the diluted solution titrated with \(5.5 \mathrm{~mL}\) of \(1 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) to react with all iodine. The \(\mathrm{I}_{2}\) from another \(25 \mathrm{~mL}\) sample was extracted after which the \(\mathrm{Fe}^{2+}\) was titrated with \(3.2 \mathrm{~mL}\) of \(1 \mathrm{MMnO}_{4}^{-}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Calculate percentage of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) in the mixture. [Hint : Reactions involved are : (1) \(\mathrm{Fe}_{3} \mathrm{O}_{4}+2 \mathrm{KI}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{FeSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}\) (2) \(\mathrm{Fe}_{2} \mathrm{O}_{3}+2 \mathrm{KI}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{FeSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+3 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}\) (3) \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \Gamma+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (4) \(\left.\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\right]\). \(25 \mathrm{~m} \mathrm{~L} \mathrm{H}_{2} \mathrm{O}_{2}\) were added to excess of acidified solution of \(\mathrm{KI}\). The iodine so liberated required \(20 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) sodium

Find out the percentage of oxalate in a given sample of an oxalate salt of which when \(0.3 \mathrm{~g}\) were dissolved in \(100 \mathrm{~mL}\) of water required \(90 \mathrm{~mL}\) of \(\mathrm{N} / 20 \mathrm{KMnO}_{4}\) solution for complete oxidation.

A sample containing \(0.4775 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) and inert materials was dissolved in water and made strongly alkaline with \(\mathrm{KOH}\) which converted \(\mathrm{NH}_{4}^{+}\) to \(\mathrm{NH}_{3}\). The liberated ammonia was distilled into exactly \(50.0 \mathrm{~mL}\) of \(0.05035 M\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The excess \(\mathrm{H}_{2} \mathrm{SO}_{4}\) was back titrated with \(11.3 \mathrm{~mL}\) of \(0.1214 \mathrm{M} \mathrm{NaOH}\). Calculate: (i) percentage of \(\mathrm{N}\), (ii) percentage of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) [At. mass of nitrogen \(=14.0078\); Mol. mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) \(\begin{array}{ll}=124.10] & \text { [M.L.N.R. 1994] }\end{array}\) [Hint : \(11.3 \mathrm{~mL} 0.1214 \mathrm{M} \mathrm{NaOH}=11.3 \mathrm{~mL} 0.1214 \mathrm{~N} \mathrm{NaOH}\) \(\equiv 11.3 \mathrm{~mL} 0.1214 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}=1.3718 \mathrm{meq} . \mathrm{H}_{2} \mathrm{SO}_{4}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) acid used for neutralisation of \(\mathrm{NH}_{3}\) $$ \begin{aligned} &=(5.035-1.3718)=3.663 \mathrm{meq} . \\ &\equiv 3.663 \mathrm{meq} . \mathrm{NH}_{3} \end{aligned} $$ Amount of ammonium oxalate \(=3.663 \times 10^{-3} \times\) Eq. mass $$ \begin{aligned} &=3.663 \times 10^{-3} \times \frac{124.10}{2} \\ &=0.2273 \mathrm{~g}] \end{aligned} $$
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