Question

Answer 1

a. \(n_{O_2}=\dfrac{12.8}{32}=0,4\left(mol\right)\)

\(n_P=\dfrac{50.22}{31}=1,62\left(mol\right)\)

Ta thấy 0,4 < 1,62 => O₂ đủ , P dư

PTHH : 4P + 5O₂ -> 2P₂O₅ 

           0,32   0,4        0,16

\(m_{P_2O_5}=0,16.142=22,72\left(g\right)\)

b. \(m_{P\left(dư\right)}=\left(1,62-0,32\right).31=40,3\left(g\right)\)

Answer 2

4P+5O₂-to>2P₂O₅

 0,32- 0,4----0,16 mol

n _O2=\(\dfrac{12,8}{32}\)=0,4 mol

n _P=\(\dfrac{50,22}{31}\)=1,62mol

=>P dư

=>m _P2O5=0,16.142=22,72g

=> m_{P dư}=1,3.31=40,3g