Answer to Question #156781 in Inorganic Chemistry for dylan

Question

95.0 grams of I₂O₅, reacts with 45.0 grams of carbon monoxide, CO, in the reaction described below. If the percentage yield is 93.0% calculate the actual yield of CO₂ in grams.

Answer 1

$I_2O_5 + 5CO \to I_2 + 5CO_2$

from the reaction above,

1 mole of I2O5 reacts with 5 moles of CO

333.8g of I2O5 reacts with 140g of CO

This means that CO is the limiting agent, because;

45.0g of CO will react with 107.3g of I2O5

from the reaction above,

5 moles of CO produces 5 moles of CO2

This means that 1 mole of CO produces 1 mole of CO2

Therefore, 28g of CO produces 44g of CO2

but since the percentage yield of CO2 is 93.0%, 40.92g of CO2 (93/100 × 44g) is produced from 1 mole of CO

rewriting,

28g of CO = 40.92g of CO2

45.0g of CO = xg

x = $\dfrac{45×40.92}{28}$ = 65.8g

Therefore, the actual yield of CO₂ is 65.8g.