Question

$P_4+5O_2⟶P_4{O}_{10}$ If 31 g of $P_4$ is reacted with excess of oxygen, the % yield is 60%, then the amount of product formed is.

Answer 1

1. The given reaction is:

$P_4+5O_2⟶P_4{O}_{10}$

2. Here, one mole of phosphorous is reacting with 5 moles of oxygen to give one mole of$P_4{O}_{10}$

3. The mass of $P_4=4\times 31=124gm$

4. The mass of $P_4{O}_{10}=4\times 31+10\times 16=284gm$

5. If the yield is 100%, then 31 gm of $P_4$ will give $=\frac{31\times 284}{124}=71gmof{P}_4{O}_{10}$

6. The percentage of yield is 60%, then the amount of product $P_4{O}_{10}$ formed is:

$\therefore Theamountofproduct=71\times 60\%$

$=71\times \frac{60}{100}$

$=42.6gof{P}_4{O}_{10}$