Question

9. Diiodine pentoxide is useful in devices such as respirators because it reacts with the dangerous gas carbon monoxide, CO, to produce relatively harmless CO2 according to the following equation: I2O5 5CO £ I2 5CO2 a. In testing a respirator, 2.00 g of carbon monoxide gas is passed through diiodine pentoxide. Upon analyzing the results, it is found that 3.17 g of I2 was produced. Calculate the percentage yield of the reaction. b. Assuming that the yield in (a) resulted because some of the CO did not react, calculate the mass of CO that passed through.

Answer 1

a)The percentage yield of the reaction is 87.3%.

b) 0.343 gram is the mass of CO that passed through.

Step-by-step explanation:

`I_2O_5+ 5CO\rightarrow  I_2 + 5CO_2`

Moles of carbon monoxide gas =
`(2.00 g)/(28 g/mol)=0.07143 mol`

According to reaction, 5 moles of carbon monoxide gives 1 moles of iodine gas,then 0.07143 moles of carbon monoxide will give :

`(1)/(5)* 0.07143 mol=0.014286 mol` of iodine gas

Mass of 0.014286 moles of iodine gas:

0.014286 mol × 254 g /mol = 3.63 g

Theoretical yield of the iodine gas = 3.63 g

Experimental yield of the iodine gas = 3.17 g

Percentage yield of the reaction :

`\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`=(3.17 g)/(3.63 g)* 100=87.3\%`

The percentage yield of the reaction is 87.3%.

b)

Mass of iodine gas produced = 3.17 g

Moles of iodine gas =
`(3.17 g)/(254 g/mol)=0.01248 mol`

According to reaction , 1 mole of iodine gas is obtained from 5 moles of carbon monoxide gas, then 0.01248 moles of iodine gas will be obtained from :

`(5)/(1)* 0.01248 mol=0.06240 mol` of carbon monoxide

Moles of carbon monoxide reacted = 0.06240 mol

Moles of carbon monoxide gas used = 0.07143 mol

Moles of carbon monoxide gas which do not reacted:

0.07143 mol - 0.06240 mol = 0.00903 mol

Mass of 0.00903 moles of carbon monoxide gas:

0.00903 mol × 28 g/mol = 0.343 g

0.343 gram is the mass of CO that passed through.