Lead forms compounds in the +2 and +4 oxidation states. All...

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: $$ \mathrm{PbX}_{4} \longrightarrow \mathrm{Pb} \mathrm{X}_{2}+\mathrm{X}_{2} $$ 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.

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:
$$
\mathrm{PbX}_{4} \longrightarrow \mathrm{Pb} \mathrm{X}_{2}+\mathrm{X}_{2}
$$
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.

Key Concepts

Conservation of Mass

The principle of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This concept is critical for balancing chemical equations and for calculating the masses of products or reactants during a reaction. In the problem, the total mass of the initial lead compound is distributed among the lead(II) halide produced and the free halogen, enabling an analysis that reveals the identity of the halogen based on the mass difference.

Atomic Mass Determination

Atomic mass determination involves comparing the experimentally observed mass ratios from a reaction with the known masses of elements to identify unknown components. Through calculation of the mass difference associated with the free halogen and equating it to the known molecular mass of a diatomic halogen, one can deduce which halogen is involved. This process links the theoretical stoichiometry with real-world atomic weights and helps in identifying the halogen based on its characteristic atomic mass.

Oxidation States

Oxidation states describe the degree of oxidation of an element in a chemical compound. In this context, lead exhibits different oxidation states (+2 and +4), which influence the chemical behavior and reactivity of its compounds. The change from a higher oxidation state to a lower one during the reaction drives the redox process, affecting how other species in the reaction, such as halide ions, are transformed.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction based on the balanced chemical equation. It allows for calculation of masses or moles of substances consumed or produced in the reaction. In this case, stoichiometric relationships between the lead(IV) halide, lead(II) halide, and free diatomic halogen are used to relate the observed masses back to the chemical formulas involved.

Transcript

00:02 The following question asks us to try and identify a halide with the given information.

00:10 So our goal is to identify x, the halide.

00:13 We're given this reaction, or a lead -4 halide, will give you a lead -2 halide and the free halide, the free halogen.

00:24 What they let us know is that our reactants, we start with 25 grams of, and the lead to halide, we produce 16 .12 grams.

00:36 Additionally, from looking at the periodic table, we know the molecular weight of lead.

00:42 So to solve this problem, we're just going to have to do a little bit of algebra.

00:47 So let's write an expression for molecular weight of pb x4.

00:59 Okay? and we don't know what the halide is, so we can't know it's molecular weight.

01:03 But what we can write is 207 .2 plus 4x.

01:13 Okay, x is going to be a number that represents the molecular weight of our unknown halide.

01:19 And so this is in units of, yeah, that'll work, grams per mole.

01:32 Okay, so that's for our reactant for the product.

01:43 I'm doing the lead to halide, but you could easily just do the halide produced, use conservation with mass, figure out how many grams here are produced, but both ways should work.

01:56 And so here it's going to be similar.

01:59 207 .2 plus this time we only have two of the halide.

02:05 And this is in grams per mole.

02:10 Again, okay.

02:13 Let's check if our equation is balanced.

02:16 So we have.

02:17 Of one lead, one lead, four halide, four halide.

02:21 Okay, so that's a balanced equation.

02:23 So we can say that every one mole of the lead four halide will produce one mole of the lead two halide.

02:32 Okay, so what will we do with that information? let's just write it out first.

02:42 X4 is going to produce one mole pbx2.

02:49 Okay, good to know.