Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. Which reactant remains unconsumed at the end of the com-bustion?

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

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

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g / 76.139 g/mol = 0.46 moles

Number of moles of O2 = 30g/15.999 g/mol = 1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2 * 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92*64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 * 0.46 moles of O2 which is = 1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting