Question

What weights of $P_4{O}_6$ and $P_4{O}_{10}$ will be produced by the combustion of $2g$ of $P_4$ in $2g$ of oxygen leaving no $P_4$ and $O_2$ ?

Answer 1

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

$2g$ $2g$

$\frac{2}{124}$ moles $\frac{2}{32}$ moles

In reaction : $P_4+3O_2\to P_4{O}_6$

moles of $O_2$ required to react with $\frac{1}{62}$ moles of $P_4$ completely

$=\frac{1}{62}\times 3=\frac{3}{62}<\frac{1}{16}$

In reaction : $P_4+5O_2\to P_4{O}_{10}$

mole of $O_2$ required to react completely with

$\frac{1}{62}$ moles of $P_4=\frac{1}{62}\times 5=\frac{5}{62}>\frac{1}{16}$

SO , $P_4$ will first react with $O_2$ to produce $P_4{O}_6$ & then the remaining $O_2$ will react with $P_4{O}_6$ by produce $P_4{O}_{10}$

$P_4+3O_2\to P_4{O}_6$ + $2O_2\to P_4{O}_{10}$

$\frac{1}{62}\to \frac{3}{62}$ moles $\frac{1}{62}$ mole $(\frac{1}{16}-\frac{3}{62})$

$Wt.$ of $P_4{O}_6$ produced = $\frac{1}{62}\times 220g=3.55g$

moles of $O_2$ (remaining ) = $\frac{1}{16}-\frac{3}{62}=\frac{62-48}{62\times 16}=0.014133$

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

$0.014$ moles

$\therefore$ moles of $P_4{O}_6$ (reacted) = $\frac{0.014}{2}=0.00706$

$\therefore$ moles of $P_4{O}_6$ (remained) = $\frac{1}{62}-0.00706=0.00907$

moles of $P_4{O}_{10}$ (produced) = $\frac{0.014113}{2}=0.00706$

$Wt.$ of $P_4{O}_{10}$ (produced) = $0.00706\times 284=2.004g$

$Wt.$ of $P_4{O}_6$ (produced) = $0.00906\times 220=1.996g$