Question

What mass of $P_2{O}_6$ will be produced by the combustion of 2.0 g of $P_4$ with 2.0 g of $O_2$?

• A. 0.0145 mol
• B. 0.072 mol
• C. 0.029 mol
• D. 0.0048 mol

Answer 1

Answer: (B)

0.072 mol

Moles of $P_4=\frac{2.0}{31\times 4}=0.016$ mol.
Moles of $O_2=\frac{2.0}{32}=0.1625$ mol.
This is not enough $O_2$ to produce $P_4{O}_{10}$, which would have required $5\times 0.016=0.08$ mol $O_2$.
Thus, $\left(0.016mol{P}_4\right)\left(\frac{3mol{O}_2}{mol{P}_4}\right)$ $=0.048mol{O}_2$ used to make $P_4{O}_6$.
Moles of $O_2$ unreacted $=(0.0625-0.048)=0.0145mol{O}_2$.
This $O_2$ can react according to the equation:
$P_4{O}_6+2O_2\to P_4{O}_{10}$
Then,
$\left(0.0415mol{O}_2\right)\left(\frac{1mol{P}_4{O}_{10}}{2mol{O}_2}\right)=0.00721mol{P}_4{O}_{10}$
Of the $0.016$ mol of $P_4{O}_6$ originally formed, $0.00725$ mol is converted to $P_4{O}_{10}$, leaving $(0.016-0.00721)=0.009$ mol $P_4{O}_6$.
(Mw of $P_4{O}_6=220)$.
Weight of $P_4{O}_6=0.009\times 220=1.98g{P}_4{O}_6$.