using the equation cs2+3o2=co2+2so2 what is the percent yield of so2 if the burning of 12.5g of cs2 produces 15.12 g of SO2

Answer;

= 71.96%

Explanation;

1 mole of CS2 = 76.133 g

1 mole of SO2 = 64.06 g

1 mole of CS2 reacts with 3 moles of Oxygen to produce 2 moles of SO2

Moles of CS2 in 12.5 g = 12.5/76.133

= 0.164 moles

Mass of SO2 = 0.164 mol * 2 * 64.06

= 21.01168 g

Therefore;

Percentage yield of SO2

= 15.12/21.01168 * 100%

= 71.96%