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: \begin{equation} \mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \end{equation} \begin{equation} \begin{array}{l}{\text { a. Balance the equation, and determine the theoretical }} \\ {\text { yield of } \mathrm{PbO} \text { if } 200.0 \mathrm{g} \text { of } \mathrm{PbS} \text { is heated. }} \\ {\text { b. What is the percent yield if } 170.0 \mathrm{g} \text { of } \mathrm{PbO} \text { is obtained? }}\end{array} \end{equation}

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

Key Concepts

Percent Yield

Percent yield is a measure of the efficiency of a reaction, defined as the ratio of the actual yield to the theoretical yield, multiplied by 100. It helps in understanding the practical efficiency of a reaction process in real conditions.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from a given amount of reactant under ideal conditions. It is calculated using stoichiometry and represents an ideal scenario without any losses.

Balancing Chemical Equations

Balancing chemical equations ensures that the number of atoms for each element is the same on both sides of the equation. This is crucial to obey the law of conservation of mass and to set up the correct stoichiometric relationships used in calculations.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It is used to calculate the amounts of substances consumed and produced, based on the balanced chemical equation.

Transcript

00:01 In this question, we are given the equation of the reaction between lead to sulfide and oxygen, which produces lead to oxide and sulfur dioxide.

00:24 First, we need to balance this equation, and then we need to determine the theoretical yield of lead to oxide when you have 200 grams.

00:38 Grams of lead to sulfide completely reacting okay so we need to determine the theoretical view here but first let's balance the equation we have one lead one lead one sulfur one sulfur two oxygen three oxygen so we need three oxygines here i don't like fractions so we're gonna multiply the core equation by two and then when we do that we get two lead sulfide the three half becomes three and then two of lead to oxide and two of sulfur dioxide so this is the balance equation now let's determine the theoretical yield of of so it is the theoretical view of lead to oxide not of sulfur oxide so let to oxide so so this is the reactor involved this is the product involved we're going to establish the relationship the mole ratio between lead to sulfide and lead to oxide they react in a two to two more ratio okay two moles of lead to oxide react with two moles i produce two moles of lead to oxide so two moles of lead to sulfide produce two modes of lead to oxide because the information is given in grants and the answer must be giving in grants we need to convert these moles to grant first, we multiply here by the modern mass of lead to sulfide.

02:56 Lead, the atomic weight of lead is 207 .2, and the atomic weight of sulfur, is 32 .06.

03:14 Here we do the same thing.

03:16 Lead is 207 .2, and oxygen is 16.

03:22 This is grams per mole...

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