Question

What weights of $P_4{O}_6$ and $P_4{O}_{10}$ will be produced by the combustion of $31\text{g}$ of $P_4$ in $32\text{g}$ of oxygen assuming complete combustion?

• A. 2.75 g, 219.5 g
• B. 27.5 g, 35.5 g
• C. 55 g, 56.77 g
• D. 17.5 g, 190.5 g

Answer 1

The correct option is B 27.5 g, 35.5 g

The equations for the given reactions is:
$P_4+3O_2\to P_4{O}_6$
$P_4+5O_2\to P_4{O}_{10}$
When we combine the above, we get
$2P_4+8O_2\to P_4{O}_6+P_4{O}_{10}$
We can calculate the mass of the individual products formed by calculating the mass of phosphorus and oxygen reacted individually.
It is important to note that the mass of $P_4$used would be the same in both cases, i.e. $\frac{31}{2}=15.5g$
Oxygen however reacts in different ratios, out of a total of 8 parts, 3 parts combine to form $P_4{O}_6$ and $P_4{O}_{10}$ is formed by the other 5 parts.
Mass of oxygen in $P_4{O}_6:$ $\frac{3\times 32g}{8}$=$12g$
Mass of oxygen in $P_4{O}_{10}$: $5\times O_2$ =$\frac{5\times 32g}{8}$=$20g$
Hence, the weights of the products would be:
$P_4{O}_6=15.5g+12g$= $27.5g$
$P_4{O}_{10}=15.5g+20g=35.5g$