Carbon disulfide burns in oxygen to yield carbon dioxide and sulfur dioxide according to the following chemical equation. CS2 (l) + 3O2 (g) → CO2 (g) + 2SO2 (g)

a. If 1.00 mol CS2 reacts with 1.00 mol O2, identify the limiting reactant.

b. How many moles of excess reactant remain?

c. How many moles of each product are formed?

a) the limiting reactant is 02

b) There will remain 0.667 moles of CS2

c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2

Explanation:

Step 1: Data given

Number of moles of CS2 = 1.00 mol

Number of moles of O2 = 1.00 mol

Molar mass of O2 = 32 g/mol

Molar mass of CS2 = 76.14 g/mol

Step 2: The balanced equation

CS2 (l) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Step 3: Calculate the limiting reactant

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

O2 is the limiting reactant. It will completely be consumed. (1.00 mol).

CS2 is in excess. There will react 1.00 / 3 = 0.333 moles

There will remain 1.00 - 0.333 = 0.667 moles of CS2

Step 4: Calculate moles of CO2 and SO2

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2

For 1.00 mol of O2 we have 1.00 / (3/2) = 0.667 moles of SO2