Question

What weights of $P_4{O}_6$ and $P_4{O}_{10}$ will be produced by the combustion of 31 g of $P_4$ in 32 g of oxygen leaving no $P_{4}and{O}_2$?

• A. 17.5 g, 190.5 g
• B. 2.75 g, 219.5 g
• C. 55 g, 71 g
• D. 27.5 g, 35.5 g

Answer 1

The correct option is D 27.5 g, 35.5 g

Calculating the moles of $P_4{O}_6$ formed:

Moles of $P_4$ reacted $=\frac{mass}{molarmass}=\frac{31}{124}=0.25$

Moles of $O_2$ reacted $=\frac{mass}{molarmass}=\frac{32}{32}=1$

Chemical reaction involved:

$P_4+3O_2\to P_4{O}_6$

Initial
moles 0.25 1

Final 0.25 - 0.25 1 - 3 × 0.25 0.25
moles =0 = 0.25

Thus, 0.25 mol of $P_4{O}_6$ will be formed along with 0.25 mol of $O_2$ still left in the mixture.

Calculating the amount of $P_4{O}_{10}$ formed

Moles of $P_4{O}_6$ present = 0.25 mol
Moles of $O_2$ present = 0.25 mol

Chemical reaction involved:

$P_4{O}_6\to 2O_2\to P_4{O}_{10}$

Initial 0.25 0.25
moles

Final 0.25 - $\frac{0.25}{2}$ 0.25 - 0.25 $\frac{0.25}{2}=0.125$
Moles = 0.125 =0

Mass of $P_4{O}_{10}$ formed
= Number of moles × molar mass
= 0.125 mol × 284 g$mo{l}^{-1}$ = 35.5 g
After the formation of $P_4{O}_{10}$,
Moles of $P_4{O}_6$ left = 0.125 mol
Mass of $P_4{O}_6$
= Number of moles × molar mass
= 0.125 mol × 220 $gmo{l}^{-1}$
= 27.5 g

Hence, 27.5 g of $P_4{O}_6$ and 35.5 g of $P_4{O}_{10}$ are produced.