# How do you balance PH_3 + O_2 -> P_4O_10 + H_2O?

Mar 5, 2016

$4 P {H}_{3} \left(g\right) + 8 {O}_{2} \left(g\right) \rightarrow {P}_{4} {O}_{10} \left(s\right) + 6 {H}_{2} O \left(l\right)$
Combustion reactions like these can usually be balanced directly, without that tedious mucking about with redox half equations. Hydrocarbon combustion reactions might be balanced the same way. Note that this IS FORMALLY a redox reaction: $P \left(- I I I\right)$ is oxidized to $P \left(+ V\right)$. What is reduced here?