Question

Question 4Calculate the maximum mass (in g) of Pb(s) that can be obtained from the reaction of 393. g PbS with 593. g PbO. Enter your answer as an integer.2 PbO(s) + PbS(s) → 3 Pb(s) + SO₂(g)826 margin of error

Answer 1

1) Write the chemical equation.

`2PbO+PbS\rightarrow3Pb+SO_2`

2) List the known and unknown quantities.

Sample: PbS

Mass: 393. g.

Sample: PbO.

Mass: 593. g.

Pb produced: unknown.

3) Convert masses to moles

3.1-Convert mass of PbS to moles of PbS.

The molar mass of PbS is 239.2650 g/mol.

`mol\text{ }PbS=393.\text{ }g\text{ }PbS*\frac{1\text{ }mol\text{ }PbS}{239.2650\text{ }g\text{ }PbS}=1.642530\text{ }mol\text{ }PbS`

We have 1.64 mol PbS

3.2-Convert mass of PbO to moles of PbO.

The molar mass of PbO is 223.1994 g/mol.

`mol\text{ }PbO=593.\text{ }g\text{ }PbO*\frac{1\text{ }mol\text{ }PbO}{223.1994\text{ }g\text{ }PbO}=2.656817\text{ }mol\text{ }PbO`

We have 2.66 mol PbO.

4) Limiting reactant

4.1-How many moles of PbO do we need to use all of the PbS?

The molar ratio between PbO and PbS is 2 mol PbO: 1 mol PbS.

`mol\text{ }PbO=1.64\text{ }mol\text{ }PbS*\frac{2\text{ }mol\text{ }PbO}{1\text{ }mol\text{ }PbS}=3.28\text{ }mol\text{ }PbO`

We need 3.28 mol PbO and we have 2.66 mol PbO. We do not have enough PbO. This is the limiting reactant.

4.2-How many moles of PbS do we need to use all of the PbO?

The molar ratio between PbO and PbS is 2 mol PbO: 1 mol PbS.

`mol\text{ }PbS=2.66\text{ }mol\text{ }PbO*\frac{1\text{ }mol\text{ }PbS}{2\text{ }mol\text{ }PbO}=1.33\text{ }mol\text{ }PbS`

We need 1.33 mol PbS and we have 1.64 mol PbS. We have enough PbS. This is the excess reactant.

5) Moles of Pb produced in the reaction from the limiting reactant.

Limiting reactant: We have 2.66 mol PbO.

The molar ratio between PbO and Pb is 2 mol PbO: 3 mol Pb.

`mol\text{ }Pb=2.66\text{ }mol\text{ }PbO*\frac{3\text{ }mol\text{ }Pb}{2\text{ }mol\text{ }PbO}=3.99\text{ }mol\text{ }Pb`

6) Convert moles of Pb to mass of Pb.

The molar mass of Pb is 207.2000 g/mol.

`g\text{ }Pb=3.99\text{ }mol\text{ }Pb*\frac{207.2000\text{ }g\text{ }Pb}{1\text{ }mol\text{ }Pb}=826.728\text{ }g\text{ }Pb`

The mass of Pb produced is 827. g Pb.

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