Obtain grams of lead

Question

Using the balanced equation 2PbS + 3O2 - 2Pb + 2SO3 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 in to the correct number of significant figures. And be sure to work with both Pbs and O2

Answer 1

This is a limiting reactant stoichiometry calculation problem because you are given the amount of more than one reactant. Remembering how to work these problem, the general method steps are:

Grams reactant → moles reactant → moles of product → grams of product

where to get moles we divide by the molar mass of the compound and to relate moles of reactant to moles of product we use the coefficients from the balanced equation

Because you have 2 reactants you will need to perform the above steps for both reactants.

The limiting reactant of any reaction which determines the amount of product for a reaction is the one which YIELDS the least amount of product (expressed in moles or grams)! The yielded amount WILL then be your answer!

2PbS + 3O₂ → 2Pb + 2SO₃

Molar masses are 239.3 for PbS; 32.0 for O₂ and 207.2 for Pb

2.54 g PbS X (1 mol PbS / 239.3 g PbS ) X (2 mol Pb/2 mol PbS) = 0.0106 mol Pb

1.88 g O₂ X (1 mol O₂ /32.00 g O₂) X (2 mol Pb /3 mol O₂ ) = 0.0392 mol Pb

Limiting reagent is PbS and the amount of Pb in moles = 0.0106 mol.

Grams of Pb = 0.0106 mol Pb X 207.2 g Pb / 1 mol Pb = 2.20 g Pb

The explanation here was given in such a way that the same steps CAN be used for ANY reaction stoichiometry problem given more than one reagent!

Hope this helps!