Question

When lead(II) sulfide ore burns in oxygen, the products are solid lead(II) oxide and sulfur dioxide gas. a. Write the balanced equation for the reaction. b. How many grams of oxygen are required to react with \(0.125\) mole 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?

Short answer

a. 2 PbS + 3 O_2 → 2 PbO + 2 SO_2. b. 6.00 grams. c. 17.4 grams. d. 137.0 grams.

Step-by-step solution

Write the unbalanced equation

Identify the reactants and products in the reaction. The reactants are lead(II) sulfide (\text{PbS}) and oxygen (\text{O}_2). The products are lead(II) oxide (\text{PbO}) and sulfur dioxide (\text{SO}_2).Unbalanced equation:\[ \text{PbS} + \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \]

Balance the chemical equation

Balance the equation by adjusting coefficients to have the same number of each type of atom on both sides.Balanced equation:\[ 2 \text{PbS} + 3 \text{O}_2 \rightarrow 2 \text{PbO} + 2 \text{SO}_2 \]

Calculate grams of oxygen required for 0.125 moles of PbS

Using the balanced equation, determine the mole ratio between PbS and O2:\[ 2 \text{PbS} : 3 \text{O}_2 \]For 0.125 moles of PbS, the moles of O2 required is:\[ 0.125 \text{ moles PbS} \times \frac{3 \text{ moles O}_2}{2 \text{ moles PbS}} = 0.1875 \text{ moles O}_2 \]Convert moles O2 to grams:\[ 0.1875 \text{ moles O}_2 \times 32 \text{ g/mole} = 6.00 \text{ grams O}_2 \]

Calculate grams of sulfur dioxide produced from 65.0 grams of PbS

First, convert grams of PbS to moles:\[ \text{Molar mass of PbS} = 239.2 \text{ g/mole} \]\[ 65.0 \text{ g PbS} \times \frac{1 \text{ mole PbS}}{239.2 \text{ g}} = 0.272 \text{ moles PbS} \]Using the mole ratio from the balanced equation, determine moles of SO2:\[ 0.272 \text{ moles PbS} \times \frac{2 \text{ moles SO}_2}{2 \text{ moles PbS}} = 0.272 \text{ moles SO}_2 \]Convert moles SO2 to grams:\[ 0.272 \text{ moles SO}_2 \times 64.07 \text{ g/mole} = 17.4 \text{ grams SO}_2 \]

Calculate grams of PbS required to produce 128 grams of PbO

First, convert grams of PbO to moles:\[ \text{Molar mass of PbO} = 223.2 \text{ g/mole} \]\[ 128 \text{ g PbO} \times \frac{1 \text{ mole PbO}}{223.2 \text{ g}} = 0.573 \text{ moles PbO} \]Using the mole ratio from the balanced equation, determine moles of PbS:\[ 0.573 \text{ moles PbO} \times \frac{2 \text{ moles PbS}}{2 \text{ moles PbO}} = 0.573 \text{ moles PbS} \]Convert moles PbS to grams:\[ 0.573 \text{ moles PbS} \times 239.2 \text{ g/mole} = 137.0 \text{ grams PbS} \]

Key concepts

balancing chemical equations

Balancing a chemical equation is essential because it ensures the law of conservation of mass is followed. This law states that mass cannot be created or destroyed in a chemical reaction. To balance an equation, we need to have the same number of each type of atom on both sides.
Begin by writing down the unbalanced equation with the reactants and products. For the reaction of lead(II) sulfide with oxygen, we start with: \[\text{PbS} + \text{O}_2 \rightarrow \text{PbO} + \text{SO}_2 \] Identify the number of atoms for each element on both sides of the equation, and adjust coefficients as needed: \[2 \text{PbS} + 3 \text{O}_2 \rightarrow 2 \text{PbO} + 2 \text{SO}_2 \] Now, each side of the equation has the same number of lead, sulfur, and oxygen atoms, so the equation is balanced.

mole-to-mole conversions

Mole-to-mole conversions use the relationships established by a balanced equation to convert between moles of reactants and products.
For example, in the reaction of PbS and O2, the balanced equation shows: \[2 \text{PbS} : 3 \text{O}_2 \] This means that for every 2 moles of PbS, 3 moles of O2 are required. If we have 0.125 moles of PbS and want to find out how many moles of O2 are needed, we set up a ratio: \[0.125 \text{ moles PbS} \times \frac{3 \text{ moles O}_2}{2 \text{ moles PbS}} = 0.1875 \text{ moles O}_2 \] This ratio derived from the balanced equation guides the calculation.

mass-to-mass conversions

Mass-to-mass conversions allow us to determine the mass of a reactant or product using their moles and molar masses.
Consider the conversion of grams of PbS to grams of SO2 using the balanced equation. First, we convert the given mass of PbS to moles: \[\text{Molar mass of PbS} = 239.2 \text{ g/mole} \]\[65.0 \text{ g PbS} \times \frac{1 \text{ mole PbS}}{239.2 \text{ g}} = 0.272 \text{ moles PbS} \] Then, use the mole-to-mole ratio to find the moles of SO2 produced: \[0.272 \text{ moles PbS} \times \frac{2 \text{ moles SO}_2}{2 \text{ moles PbS}} = 0.272 \text{ moles SO}_2 \] Finally, convert the moles of SO2 to grams: \[0.272 \text{ moles SO}_2 \times 64.07 \text{ g/mole} = 17.4 \text{ grams SO}_2 \] These steps provide a systematic way to solve mass-to-mass conversion problems.

molar mass calculations

Molar mass calculations are crucial in stoichiometry to convert between grams and moles.
The molar mass of a compound is the sum of the atomic masses of all atoms in the molecule. For lead(II) sulfide (PbS), the molar mass is: \[\text{Molar mass of PbS} = \text{Atomic mass of Pb} + \text{Atomic mass of S} \]\[ \text{Molar mass of PbS} = 207.2 \text{ g/mole} + 32.1 \text{ g/mole} = 239.2 \text{ g/mole} \] Similarly, the molar mass of sulfur dioxide (SO2) is calculated as follows: \[ \text{Molar mass of SO}_2 = \text{Atomic mass of S} + 2 \times \text{Atomic mass of O} \]\[ \text{Molar mass of SO}_2 = 32.1 \text{ g/mole} + 2 \times 16.00 \text{ g/mole} = 64.07 \text{ g/mole} \] Knowing these molar masses enables accurate calculations for stoichiometric problems.