Question

The phophorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by nurning phosphorus in oxygen.

Answer 1

Answer : The percent yield of $P_4O_{10}$ is, 87.7 %

Solution : Given,

Moles of $P_4$ = 0.200 mole

Moles of $O_2$ = 0.200 mole

Molar mass of $P_4O_{10}$ = 283.9 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$P_4+5O_2\rightarrow P_4O_{10}$

From the balanced reaction we conclude that

As, 5 mole of $O_2$ react with 1 mole of $P_4$

So, 0.200 moles of $O_2$ react with $\frac{0.200}{5}=0.04$ moles of $P_4$

From this we conclude that, $P_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $P_4O_{10}$

From the reaction, we conclude that

As, 5 mole of $O_2$ react to give 1 mole of $P_4O_{10}$

So, 0.200 moles of $O_2$ react to give $\frac{0.200}{5}=0.04$ moles of $P_4O_{10}$

Now we have to calculate the mass of $P_4O_{10}$

$\text{ Mass of }P_4O_{10}=\text{ Moles of }P_4O_{10}\times \text{ Molar mass of }P_4O_{10}$

$\text{ Mass of }P_4O_{10}=(0.04moles)\times (283.9g/mole)=11.4g$

Theoretical yield of $P_4O_{10}$ = 11.4 g

Experimental yield of $P_4O_{10}$ = 10.0 g

Now we have to calculate the percent yield of $P_4O_{10}$

$\% \text{ yield of }P_4O_{10}=\frac{\text{ Experimental yield of }P_4O_{10}}{\text{ Theretical yield of }P_4O_{10}}\times 100$

$\% \text{ yield of }P_4O_{10}=\frac{10.0g}{11.4g}\times 100=87.7\%$

Therefore, the percent yield of $P_4O_{10}$ is, 87.7 %