a)\(4P+5O2-->2P2O5\)
\(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_P\left(\frac{0,2}{4}\right)< n_{O2}\left(\frac{0,3}{5}\right)=>O2dư\)
\(n_{O2}=\frac{5}{4}n_P=0,25\left(mol\right)\)
\(n_{O2}dư=0,3-0,25=0,05\left(mol\right)\)
\(m_{O2}dư=0,05.32=1,6\left(g\right)\)
b)\(n_{P2O5}=\frac{1}{2}n_P=0,1\left(mol\right)\)
\(m_{P2O5}=0,1.142=14,2\left(g\right)\)
\(n_{P\left(đb\right)}=\frac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2\left(đb\right)}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(4P+5O_2\rightarrow2P_2O_5\)
\(\frac{n_{P\left(đb\right)}}{n_{P\left(PTHH\right)}}\) \(\frac{n_{O_2\left(đb\right)}}{n_{O_2\left(PTHH\right)}}\)
\(\frac{0,2}{4}< \frac{0,3}{5}\)
\(\Rightarrow\) P hết, O2 dư
Theo PTHH:\(n_{O_2}=\frac{5}{4}n_P=\frac{5}{4}.0,2=0,25\left(mol\right)\)
\(n_{P_2O_5}=\frac{1}{2}n_P=\frac{1}{2}.0,2=0,1\left(mol\right)\)
\(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
\(n_{O_2\left(dư\right)}=n_{O_2\left(đb\right)}-n_{O_2\left(PTHH\right)}=0,3-0,25=0,05\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)