a/ Ta có: \(n_{O_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
PTHH:
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 5
x 0.2
\(=>x=\dfrac{4\cdot0.2}{5}=0.16=n_P\)
\(=>m_P=0.16\cdot31=4.96\left(g\right)\) hay a=4.96
b/ PTHH:
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 2
0.16 y
\(=>y=\dfrac{0.16\cdot2}{4}=0.08=n_{P_2O_5}\)
\(=>m_{P_2O_5}=0.08\cdot\left(31\cdot2+16\cdot5\right)=11.36\left(g\right)\)