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

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 $6.023\times 10^{23}$ 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:

$\text{Number of moles of phosphorous}=\frac{3.07g}{31g/mol}=0.1moles$

given mass of oxygen $O_2$ = 6.09 g

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

Putting in the values we get:

$\text{Number of moles of oxygen}=\frac{6.09g}{32g/mol}=0.2moles$

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 =$\frac{5}{4}\times 0.1=0.125$ 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 =$\frac{2}{4}\times 0.1=0.05$ moles of $P_2O_5$

mass of $P_2O_5=moles\times {\text {Molar mass}}=0.05\times 284=14.2g$

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