Question

O2 + 2C0-2CO2

4.62 grams of oxygen and 2.98 grams of carbon monoxide are placed in a closed

reaction vessel and the mixture is ignited. Combustion occurs until one of the

gases is totally consumed.

1. Identify the limiting reagent and the excess reagent

2. Determine the amount of excess reagent.

3. Calculate the grams of carbon dioxide produced.

Answer 1

Answer: 1.
`CO` is the limiting reagent and
`O_2` is the excess reagent.

2. 2.91 g of excess reagent

3. 4.66 g of carbon dioxide produced.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} CO=(2.98g)/(28g/mol)=0.106moles`

`\text{Moles of} O_2=(4.62g)/(32g/mol)=0.144moles`

`O_2+2CO\rightarrow 2CO_2`

According to stoichiometry :

2 moles of
`CO` require = 1 mole of
`O_2`

Thus 0.106 moles of
`CO` will require=
`(1)/(2)* 0.106=0.053moles` of
`O_2`

Thus
`CO` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

Amount of excess reagent = ( 0.144-0.053) moles = 0.091 moles

Mass of excess reagent
`O_2=moles* {\text {Molar mass}}=0.091moles* 32g/mol=2.91g`

As 2 moles of
`CO` give = 2 moles of
`CO_2`

Thus 0.106 moles of
`CO` give =
`(2)/(2)* 0.106=0.106moles` of
`CO_2`

Mass of
`CO_2=moles* {\text {Molar mass}}=0.106moles* 44g/mol=4.66g`