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Chapter 20: Problem 41

The compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) ?

Short Answer

Expert verified
The mole ratio of lead(II) to lead(IV) in the compound $\mathrm{Pb}_{3} \mathrm{O}_{4}$ (red lead) is 2:1.

Step by step solution

01

Determine the oxidation number of Pb in the compound

First, we need to determine the oxidation number of each lead atom in the compound. It is given that red lead contains a mixture of lead(II) and lead(IV) oxidation states. Let x represent the number of moles of lead(II) with oxidation state +2 and y represent the number of moles of lead(IV) with oxidation state +4. There are a total of 3 moles of lead in Pb3O4, so we have: x + y = 3.
02

Determine the total charge in the compound

Next, we need to determine the total charge in the compound. The total charge of the compound is the sum of the charges of all the elements in the compound. Oxygen has an oxidation state of -2. Since there are four oxygen atoms in the compound, the total charge due to oxygen is -8. The total charge due to lead(II) is +2x and due to lead(IV) is +4y. The total charge in the compound will be 0 since it is a neutral compound. Combining all these charges, we get the equation: 2x + 4y - 8 = 0
03

Solve the equations for x and y

We now have two equations: 1) x + y = 3 2) 2x + 4y - 8 = 0 We solve these simultaneous equations to find the values of x and y. From equation 1, we can express x as x = 3 - y. Substituting this into equation 2, we get: 2(3 - y) + 4y - 8 = 0 Expanding and simplifying the equation, we obtain: 6 - 2y + 4y - 8 = 0 2y - 2 = 0 Now, divide by 2: y = 1 Now, substitute y back into equation 1 to find x: x + 1 = 3 x = 2
04

Calculate the mole ratio of lead(II) to lead(IV)

Now that we know the number of moles of lead(II) (x = 2) and lead(IV) (y = 1) in the compound, we can determine the mole ratio of lead(II) to lead(IV). The mole ratio is given by x:y, which is 2:1 in this case. Therefore, the mole ratio of lead(II) to lead(IV) in the compound Pb3O4 is 2:1.

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Most popular questions from this chapter

Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}\), is used to soften hard water by removing calcium ions from hard water through the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \rightarrow\) Although \(\mathrm{CaCO}_{3}(s)\) is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of \(\mathrm{CaCO}_{3}\) in water \(\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right)\)

The oxyanion of nitrogen in which it has the highest oxidation state is the nitrate ion \(\left(\mathrm{NO}_{3}^{-}\right)\). The corresponding oxyanion of phosphorus is \(\mathrm{PO}_{4}^{3-}\). The \(\mathrm{NO}_{4}{ }^{3-}\) ion is known but not very stable. The \(\mathrm{PO}_{3}^{-}\) ion is not known. Account for these differences in terms of the bonding in the four anions.

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in $$: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}$$ be described in terms of orbitals?

Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: Suppose \(25.00 \mathrm{~g}\) of a lead(IV) halide reacts to form \(16.12 \mathrm{~g}\) of a lead(II) halide and the free halogen. Identify the halogen.

Complete and balance each of the following reactions. a. the reaction between sulfur dioxide gas and oxygen gas b. the reaction between sulfur trioxide gas and water c. the reaction between concentrated sulfuric acid and sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\)
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