Question

Question #168e9

Answer 1

`6.49xx10^(-2)mol(O_2)`

Explanation:

The reaction is:
`P_4(s) + 3O_2(g) -> P_4O_6(s)`

We should calculate first the number of mole of `P_4O_6(s)`:
`n=m/(MM) =(4.76cancel(g))/(219.88cancel(g)/(mol))=2.16xx10^(-2) mol`

The number of mole of oxygen needed to produce `4.76g` of `P_4O_6(s)`:

`n_(O_2)=2.16xx10^(-2)cancel(mol(P_4O_6))xx(3 mol(O_2))/(1 cancel(mol(P_4O_6)))`
`=6.49xx10^(-2)mol(O_2)`