Question

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.88 g of O2? Express your answer to the correct number of significant figures and you must show all work in the form of dimensional analysis as shown in Lesson 4. (Hint: be sure to work the problem with both PbS and O2).

Answer 1

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams