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
1 answer:
<h3><u>Answer;</u></h3>
= 71.96%
<h3><u>Explanation</u>;</h3>
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%
<u> = 71.96%</u>
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