a. \(n_P=\dfrac{3.1}{31}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{3.2}{32}=0,1\left(mol\right)\)
PTHH : 4P + 5O2 -to> 2P2O5
0,1 0,04
Xét tỉ lệ : \(\dfrac{0.1}{4}>\dfrac{0,1}{5}\) => P dư , O2 đủ
\(m_{P_2O_5}=0,04.142=5,68\left(g\right)\)
b. \(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
4P+5O2-to>2P2O5
0,08--0,1-------0,04
n P=\(\dfrac{3,1}{31}\)=0,1 mol
n O2=\(\dfrac{3,2}{32}\)=0,1 mol
=>P dư
=>m P2O5=0,04.142=5,68g
=>VO2=0,1.22,4=2,24l