Answer: 2.24 grams of Pb
Explanation:
<u>Step 1</u>
Balanced chemical reaction;
2PbS + 3O2 → 2Pb + 2SO3
<u>Step 2</u>
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
<u>Step 3</u>
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.
<u>Step 4</u>
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
<u>Step 5</u>
Mass of Pb
Mass = moles * molar mass
Mass of Pb =0.0108 moles * 207.2 g/mol
Mass of Pb = 2.24 grams
