PTHH:
a) \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Vì \(\dfrac{n_P}{4}=\dfrac{0,2}{4}=0,05>\dfrac{n_{O_2}}{5}=\dfrac{0,15}{5}=0,03\)
=> P dư
Theo PTHH: \(n_{P_{2\left(p.ứ\right)}}=\dfrac{4}{5}n_{O_2}=0,12\left(mol\right)\)
\(\rightarrow n_{P\left(dư\right)}=0,2-0,12=0,08\left(mol\right)\)
\(\rightarrow m_{P\left(dư\right)}=0,08.31=2,48\left(g\right)\)
c)
Theo PTHH: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,06\left(mol\right)\)
\(\rightarrow m_{P_2O_5}=0,06.\left(31.2+16.5\right)=8,52\left(g\right)\)
Vậy...
a) PTHH:
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 5 2 (mol)
0,2 0,15 ? (mol)
b) Giải
\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0.2\left(mol\right)\)
\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH ta có:
\(\dfrac{0,2}{4}>\dfrac{0,15}{5}\Rightarrow\) P dư
c)
\(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{2}{5}\cdot0,15=0,06\left(mol\right)\)
\(m_{P_2O_5}=n\cdot M=0,06\cdot142=8,52\left(g\right)\)
Chúc bạn học tốt ! 
TDVN2005
a) \(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
b) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(\left\{{}\begin{matrix}\dfrac{n_P}{4}=\dfrac{0,2}{4}=0,05\\\dfrac{n_{O_2}}{5}=\dfrac{0,15}{5}=0,03\end{matrix}\right.\) \(\Rightarrow\) P dư. O2 hết như vậy tính toán theo \(n_{O_2}\)
\(\Rightarrow n_{P\left(dư\right)}=n_{P\left(bđ\right)}-n_{P\left(pư\right)}=0,2-0,12=0,08\left(mol\right)\)
\(\Rightarrow m_{P\left(dư\right)}=0,08.31=2,48\left(l\right)\)
c) Theo PTHH: \(n_{O_2}:n_{P_2O_5}=5:2\)
\(\Rightarrow n_{P_2O_5}=n_{O_2}.\dfrac{2}{5}=0,15.\dfrac{2}{5}=0,06\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,06.142=8,52\left(g\right)\)