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Chapter 22: Problem 51

A \(1.100 \mathrm{g}\) sample of copper ore is dissolved, and the \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is treated with excess KI. The liberated \(\mathrm{I}_{3}^{-}\) requires \(12.12 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) for its titration. What is the mass percent copper in the ore?

Short Answer

Expert verified
The mass percent of copper in the ore is 3.50%.

Step by step solution

01

Write balanced chemical equations

But before we proceed, the balanced equations for the reactions are: \(\mathrm{Cu}^{2+} + 2\mathrm{I}^{-} \rightarrow \mathrm{CuI}_{2}\) and \(\mathrm{I}_{3}^{-} + 2\mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow 3\mathrm{I}^{-} + 2\mathrm{S}_{4}\mathrm{O}_{6}^{2-}\). Now, from the second equation, we see that one mole of I3- consumes 2 mol of \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\).
02

Calculate moles of Na2S2O3

We first need to calculate the moles of the titrating agent, \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\). Moles can be calculated using the formula \(Moles = Molarity \times Volume\) where volume is in liters. Converting the given volume from mL to L, we get \(12.12 \mathrm{mL} = 0.01212 \mathrm{L}\). Thus, the moles of \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\) is \(0.1000 M \times 0.01212 L = 0.001212 mol\).
03

Calculate moles of I3-

From the stoichiometry of the equation in Step 1, we know 1 mol of I3- reacts with 2 mol of \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\). Therefore, the moles of I3-, which was produced from the reaction with copper is \(0.001212 mol \div 2 = 0.000606 mol\).
04

Calculate moles of Cu2+

Now, again based on the stoichiometry of the reaction, 1 mol of I3- was produced for each mole of Cu2+ that reacted. Thus, the moles of \(\mathrm{Cu}^{2+}\) is also \(0.000606 mol\).
05

Calculate mass of Copper (Cu)

The molar mass of copper (Cu) is 63.55g/mol. So, the mass of copper can be calculated as \(Mass = moles \times Molar mass\). Therefore, the mass of copper in the ore is \(0.000606 mol \times 63.55 g/mol = 0.0385 g\).
06

Calculate mass percent of Copper

The mass percent of copper in the ore can be calculated as \((Mass of Cu / Total mass of the ore) \times 100\)%. Substituting the values we get \((0.0385 g / 1.100 g) \times 100 = 3.50 \% \)

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Most popular questions from this chapter

Supply an appropriate name for each of the following: (a) \(\mathrm{HPO}_{4}^{2-} ;\) (b) \(\mathrm{Ca}_{2} \mathrm{P}_{2} \mathrm{O}_{7} ;\) (c) \(\mathrm{H}_{6} \mathrm{P}_{4} \mathrm{O}_{13}\).

The oxides of the phosphorus(III), antimony(III), and bismuth(III) are \(\mathrm{P}_{4} \mathrm{O}_{6}, \mathrm{Sb}_{4} \mathrm{O}_{6},\) and \(\mathrm{Bi}_{2} \mathrm{O}_{3} .\) Only one of these oxides is amphoteric. Which one? Which of these oxides is most acidic? Which is most basic?

Use information from this chapter and previous chapters to write plausible chemical equations to represent the following: (a) the reaction of silver metal with \(\mathrm{HNO}_{3}(\mathrm{aq})\) (b) the complete combustion of the rocket fuel, unsymmetrical dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\) (c) the preparation of sodium triphosphate by heating a mixture of sodium dihydrogen phosphate and sodium hydrogen phosphate.

In \(1968,\) before pollution controls were introduced, over 75 billion gallons of gasoline were used in the United States as a motor fuel. Assume an emission of oxides of nitrogen of 5 grams per vehicle mile and an average mileage of 15 miles per gallon of gasoline. How many kilograms of nitrogen oxides were released into the atmosphere in the United States in \(1968 ?\)

Write equations to show how to prepare \(\mathrm{H}_{2}(\mathrm{g})\) from each of the following substances: \((a) \mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{HI}(\mathrm{aq})\) (c) \(\mathrm{Mg}(\mathrm{s}) ;\) (d) \(\mathrm{CO}(\mathrm{g})\). Use other common laboratory reactants as necessary, that is, water, acids or bases, metals, and so on.
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