Lead(II) oxide is obtained by roasting galena, lead(II)...

Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is: $$\mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})$$ a. Balance the equation, and determine the theoretical yield of PbO if 200.0 g of PbS is heated. b. What is the percent yield if 170.0 g of PbO is obtained?

Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is:
$$\mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})$$

a. Balance the equation, and determine the theoretical yield of PbO if 200.0 g of PbS is heated.
b. What is the percent yield if 170.0 g of PbO is obtained?

Key Concepts

Balanced Chemical Equations

Balanced chemical equations are fundamental to chemistry, ensuring that the law of conservation of mass is observed. They provide the stoichiometric coefficients that indicate the mole ratios of reactants and products involved in the reaction. This balanced representation is essential for correctly performing calculations in chemical reactions such as yield determination and reaction scaling.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. By using the molar ratios derived from a balanced equation, chemists can calculate the amounts of substances consumed or produced, which is critical for laboratory experiments, industrial synthesis, and determining theoretical yields.

Theoretical Yield

The theoretical yield refers to the maximum amount of product that could be formed from a given quantity of reactants, as predicted by stoichiometric calculations based on a balanced chemical equation. It represents an ideal scenario in which every mole of reactant is completely converted to product without any loss, side reactions, or experimental inefficiencies.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, defined as the ratio of the actual yield to the theoretical yield multiplied by 100. It provides insight into the practical outcomes of reactions, accounting for real-world variables such as incomplete reactions, side reactions, or losses during product recovery.

Transcript

00:01 So to solve this question, the first thing we need to do is balance this unbalanced equation.

00:06 And to balance any equation, it's always good to start with oxygens.

00:11 And we compare both sides.

00:13 So on the left side, we see two oxygens.

00:16 And on the right side, we see three.

00:19 So you always look at the compound that has an odd number of oxygens and you make it even.

00:27 And we do that by increasing this amount, the amount of this compound by two.

00:42 And then, so we added two lead oxides.

00:47 But now we see that there are two leads on this side on the right side, but only one lead on the left.

00:53 So we go ahead and we write two on the left side for lead sulfide.

01:00 But now we see that there are two sulfurs.

01:04 So we go over to the right, we see so2, and we write two here as well.

01:13 So lead and sulfur are now even, but now we change the number of oxygens.

01:18 We multiplied so2 by two, and there are two oxygens in so2, but now there are four.

01:27 So we look at lead oxide, we see there are two oxygens here now.

01:32 So we add them together, 4 plus 2 is 6.

01:35 There are 6 oxygens on the right side.

01:39 So on the left side, oxygen, there are only 2 here, so we multiply that by 3, because 3 times 2 is 6.

01:48 So now this equation is balanced.

01:56 That's the first bit.

01:58 First part of a.

01:59 Now the second part of a, second part of a, we need to find the theoretical yield of pbro.

02:17 Of lead oxide, if there are 200 grams of lead sulfide being heated.

02:36 I'll just put used.

02:38 So in order to find the theoretical yield, the first thing we need to do is convert this amount into moles.

02:46 So we do that through dimensional analysis.

02:49 So we have 200 grams pvs.

02:56 You multiply this by the molar mass.

02:58 So one mole of pbs, is 239 grams per mole.

03:10 So then we cross out like units and you calculate that and we get about 0 .8.

03:28 I'll put 8357, pbs.

03:38 Now, in order to find the amount of pbo, we have to multiply this by the ratio of pbs to pbo...

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