Question

1. What is the theoretical yield of both products of the reaction between 17.56 g of carbon disulfide and an excess of oxygen? First write a balanced equation of this reaction that results in the formation of carbon dioxide and sulfur dioxide. What is the percent yield if this reaction produces 15.75 g of sulfur dioxide? Which is the limiting reagent if we only use 25.45 g of Oxygen?

Answer 1

Sol :-

Balanced chemical reaction is :

CS₂ + 3O₂ ---------------> CO₂ + 2SO₂

Number of moles of CS₂ = Given mass of CS₂ in g / Gram molar mass of CS₂​​​​​​​

= 17.56 g / 76.139 g/mol

= 0.2306 mol

Similarly,

Number of moles of O₂ = Given mass of O₂​​​​​​​ in g / Gram molar mass of O₂​​​​​​​

= 25.45 g / 32.00 g/mol

= 0.7953 mol

Because, (given mole/stoichiometric coefficient) value of CS₂ is less (that is 0.2306/1 = 0.2306) than O₂ which is 0.7953/3 = 0.2651, therefore CS₂ is limiting reagent as it is completely consumed in the chemical reaction.

Hence, CS2 is the limiting reagent.

Now,

From balanced chemical reaction :

1 mol of CS₂ gives = 1 mol of CO₂

So,

0.2306 mol of CS₂ gives = 0.2306 mol of CO₂

Now,

Mass of CO₂ produced = Moles x Gram molar mass of CO₂

= 0.2306 mol x 44 g/mol

= 10.15 g

Hence, theoretical yield of CO2 = 10.15 g

Similarly,

From balanced chemical reaction :

1 mol of CS₂ gives = 2 mol of SO₂

So,

0.2306 mol of CS₂ gives = 2 x 0.2306 mol

= 0.4612 mol of SO₂

Now,

Mass of SO₂ produced = Moles x Gram molar mass of SO₂

= 0.4612 mol x 64.066 g/mol

= 29.55 g

Hence, theoretical yield of CO2 = 29.55 g

Now,

Percent yield of SO₂ = Actual yield x 100% / Theoretical yield

= 15.75 g x 100% / 29.55 g

= 53.3%

Hence, percent yield of SO2 = 53.3%