Answer to Question #136978 in General Chemistry for Miça

Question

Consider the following balanced reaction

2PbS + 3O2—> 2PbO + 2SO2

rank the following combinations of reactants based on the maximum amount of products that can be produced from the reaction

24.18g lead sulphide and 3.38g oxygen gas

18.97g lead sulphide and 4.02g oxygen gas

22.17g lead sulphide and 3.75g oxygen gas

Answer 1

Solution:

Number of moles of a substance in a sample is the mass in g divided by the molar mass, which gives the amount in moles.

n = m/M

molar mass of PbS 239.2 gram/mol

molar mass of O₂ 32.0 gram/mol

According to the reaction equation for 2 mol of PbS (n_r PbS) are needs 3 mol of O₂ (n_r O₂) and are forms 2 mol PbO (n_r PbO) and 2 mol SO₂ (n_r SO₂).

A.

n PbS = 24.18 g PbS/239.2 g/mol PbS = 0.101 mol PbS

n O₂ = 3.38 g O₂/32 g/mol O₂ = 0.106 mol O₂

for 0.106 mol O₂ are needs n PbS = 0.106 mol O₂ × 2 mol PbS / 3 mol O₂ = 0.071 mol PbS.

the limited reactant – O₂.

And than, amount of products can be produced

n PbO = n O₂ × n_r PbO / n_r O₂

n SO₂ = n O₂ × n_r SO₂ / n_r O₂

n PbO = 0.106 mol O₂ × 2 mol PbO / 3 mol O₂ = 0.106 mol PbO

n SO₂ = 0.106 mol O₂ × 2 mol SO₂ / 3 mol O₂ = 0.106 mol SO₂

B.

n PbS = 18.97g PbS/239.2g/mol PbS = 0.079 mol PbS

n O₂ = 4.02g O₂/32g/mol O₂ = 0.126 mol

for 0.079 mol PbS are needs n O₂ = 0.079 mol PbS × 3 mol O₂ / 2 mol PbS = 0.119 mol O₂.

the limited reactant – PbS.

And than, amount of products can be produced

n PbO = n PbS × n_r PbO / n_r PbS

n SO₂ = n PbS × n_r SO₂ / n_r PbS

n PbO = 0.079 mol PbS × 2 mol PbO / 2 mol PbS = 0.079 mol PbO

n SO₂ = 0.079 mol PbS × 2 mol SO₂ / 3 mol PbS = 0.079 mol SO₂

C.

n PbS = 22.17 g PbS/239.2 g/mol PbS = 0.093 mol PbS

n O₂ = 3.75 g O₂/32 g/mol O₂ = 0.117 mol O₂

for 0.117 mol O₂ are needs n PbS = 0.117 mol O₂ × 2 mol PbS / 3 mol O₂ = 0.078 mol PbS.

the limited reactant – O₂.

And than, amount of products can be produced

n PbO = n O₂ × n_r PbO / n_r O₂

n SO₂ = n O₂ × n_r SO₂ / n_r O₂

n PbO = 0.117 mol O₂ × 2 mol PbO / 3 mol O₂ = 0.078 mol PbO

n SO₂ = 0.117 mol O₂ × 2 mol SO₂ / 3 mol O₂ = 0.078 mol SO₂

Answer: order of the combinations of reactants: 24.18g lead sulphide and 3.38g oxygen gas; 18.97g lead sulphide and 4.02g oxygen gas; 22.17g lead sulphide and 3.75g oxygen gas.