Question

Can someone help me please and clearly show step by step how to get the answer. I can't figure it out.

*For each of the following reactions, 16.0 g of each reactant is present initially.

CS₂(g)+3O₂(g)→CO₂(g)+2SO₂(g)

-Calculate the grams of product in parentheses that would be produced.
(CO₂)

Answer 1

Solution: (Grams of CO2 = 7.33 g)

The theoretical mass of CO2 is calculated by the number of moles of limiting reagent. A limiting reagent is the reagent whose number of moles is minimum.

For the given reaction,

CS2 + 3O2 = CO2 + 2SO2

Mass of CS2 = 16 g

Molar mass of CS2 = 76.139 g / mol

Thus,

Number of moles of CS2 = mass / molar mass

= 16 g / 76.139 g mol-1 = 0.210 mol

Mass of O2 = 16 g

Molar mass of O2 = 32 g /mol

Number of moles = 16 g / 32 g mol-1 = 0.50 mol

From the above reaction, it can be seen that 1 mol of CS2 requires 3 mol of O2

Hence, 0.210 mol of CS2 requires = 0.210 x 3 = 0.630 mol O2

But amount of O2 = 0.50 mol

Hence, O2 will be the limiting agent.

Since, 3 mol of O2 reacts to form 1 mol of CO2

Thus,

0.50 mol of O2 will forms = 0.50 mol x /3 = 0.167 mol CO2

Hence, number of moles of CO2 = 0.167 mol

Molar mass of CO2 = 44 g /mol

Theoretical mass of CO2 = number of moles x molar mass of

= 0.167 mol x 44  g/mol = 7.33 g