When solid lead(II) sulfide reacts with oxygen gas, the...

When solid lead(II) sulfide reacts with oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas. a. Write the balanced chemical equation for the reaction. b. How many grams of oxygen are required to react with $29.9 \mathrm{~g}$ of lead(II) sulfide? c. How many grams of sulfur dioxide can be produced when $65.0 \mathrm{~g}$ of lead(II) sulfide reacts? d. How many grams of lead(II) sulfide are used to produce $128 \mathrm{~g}$ of lead(II) oxide?

When solid lead(II) sulfide reacts with oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas.
a. Write the balanced chemical equation for the reaction.
b. How many grams of oxygen are required to react with $29.9 \mathrm{~g}$ of lead(II) sulfide?
c. How many grams of sulfur dioxide can be produced when $65.0 \mathrm{~g}$ of lead(II) sulfide reacts?
d. How many grams of lead(II) sulfide are used to produce $128 \mathrm{~g}$ of lead(II) oxide?

Key Concepts

Mole Concept

The mole is a fundamental unit in chemistry that links the mass of a substance to the number of particles it contains. It provides a bridge between the atomic scale and the macroscopic scale by using Avogadro's number, allowing chemists to convert between grams and moles during chemical calculations.

Molar Mass

Molar mass is the mass of one mole of a given substance, expressed in grams per mole. It is essential for converting between the mass of a substance and the number of moles, enabling the application of mole ratios from balanced equations to determine the amounts of reactants or products in a reaction.

Balanced Chemical Equations

Balanced chemical equations are used to represent chemical reactions in accordance with the law of conservation of mass. They provide the mole ratios among reactants and products, ensuring that the same number of atoms for each element is present on both sides of the equation, which is crucial for performing accurate stoichiometric calculations.

Stoichiometric Calculations

Stoichiometry involves using the mole ratios provided in a balanced chemical equation to calculate the quantitative relationships between reactants and products. These calculations typically include converting masses to moles, using the mole ratios to find the moles of another substance, and converting back to grams when necessary.

Transcript

00:01 For this question, you're asked to calculate the mass of a particular reactant or product, given the mass of another reactant or product.

00:08 First, you need to write the balanced chemical reaction.

00:11 Solid, lead, two sulfide reacts with oxygen.

00:15 In the process, it says it produces solid lead to oxide and sulfur dioxide.

00:22 So, lead to sulfide, lead with a 2 plus charge, sulfide with a 1 minus charge, gives us one of each of them, reacting with oxygen gas, oxygen, is diatomic, so it's o2, produces lead two oxide.

00:37 Lead 2 has a 2 plus charge, oxygen, oxide has a 2 minus charge, so we have just one of each of them, and sulfur dioxide, meaning one sulfur and dye meaning two, two oxygens.

00:52 To balance this, if we put a three halves here, we have one lead, one lead, one sulfur, one, two, three oxygens.

01:03 So how do we get three out of two? we multiply it by three halves.

01:06 Now we have three oxygens, but we can't leave it like that, so we need to multiply everything through by two in order to get our balanced chemical reaction.

01:15 Next, it asks you to calculate how many grams of oxygen are required to react with 29 .9 grams lead two sulfide.

01:24 So we'll start with our 29 .9 grams lead two sulfide.

01:28 Convert the lead two sulfide into moles lead two sulfide by dividing by the molar mass lead two sulfide 239 .3 grams per mole.

01:38 Then when we have moles lead to sulfide, we can convert the moles lead to sulfide into moles of oxygen, seeing the two to three relationship.

01:48 Then when we have moles oxygen, we simply multiply by the molar mass of oxygen to get grams of oxygen.

01:57 6 .00 grams of oxygen are required to react with 29 .9 grams led to sulfide...

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