\(n_P=\dfrac{m}{M}=\dfrac{31}{31}=1\left(mol\right)\\ n_{O_2\left(dktc\right)}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
ti lệ: 4 : 5 : 2
n(mol) 1 0,5
n(mol p/ư): 0,4<--0,5------>0,2
\(\dfrac{n_P}{4}>\dfrac{n_{O_2}}{5}\left(\dfrac{1}{4}>\dfrac{0,5}{5}\right)\)
`=>` `O_2` hết, `P` dư, tính theo`O_2`
\(n_{P\left(dư\right)}=1-0,4=0,6\left(mol\right)\)
\(m_{P\left(dư\right)}=n\cdot M=0,6\cdot31=18,6\left(g\right)\\ m_{P_2O_5}=n\cdot M=0,2\cdot\left(31\cdot2+16\cdot5\right)=28,4\left(g\right)\)