Question

Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation

4 P + 5 O2 ----> 2 P2O5

Answer 1

Answer: 14.2 grams

Explanation:-

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

given mass of phosphorous (P) = 3.07 g

Molar mass of phosphorous (P) = 31 g/mol

Putting in the values we get:

`\text{Number of moles of phosphorous}=(3.07g)/(31g/mol)=0.1moles`

given mass of oxygen
`O_2` = 6.09 g

Molar mass of oxygen
`O_2` = 32 g/mol

Putting in the values we get:

`\text{Number of moles of oxygen}=(6.09g)/(32g/mol)=0.2moles`

According to stoichiometry:

`4P+5O_2\rightarrow 2P_2O_5`

4 moles of phosphorous combine with 5 moles of oxygen

Thus 0.1 moles of phosphorous combine with =
`(5)/(4)* 0.1=0.125` moles of oxygen

Thus phosphorous acts as limiting reagent as it limits the formation of product and oxygen is the excess reagent.

4 moles of phosphorous gives= 2 moles of
`P_2O_5`

Thus 0.1 moles of phosphorous gives =
`(2)/(4)* 0.1=0.05` moles of
`P_2O_5`

mass of
`P_2O_5=moles* {\text {Molar mass}}=0.05* 284=14.2g`

Thus the theoretical yield of
`P_2O_5` is 14.2 grams.