\(n_{O2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(4P+5O_2\underrightarrow{t^o}2P_2O_5|\)
4 5 2
0,16 0,2
\(n_P=\dfrac{0,2.4}{5}=0,16\left(mol\right)\)
⇒ \(m_P=0,16.31=4,96\left(g\right)\)
Chúc bạn học tốt
\(4P+5O_2\rightarrow2P_2O_5\)
\(n_{O_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(\Leftrightarrow n_{P_2O_5}=0.5\left(mol\right)\)
\(\Leftrightarrow n_P=1\left(mol\right)\)
\(m_P=1\cdot31=31\left(g\right)\)
4P+5O2-to>2P2O5
0,16--0,2 mol
n O2=\(\dfrac{4,48}{22,4}=0,2mol\)
=> m P=0,16.31=4,96g
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)
\(m_P=n.M=\left(\dfrac{0,2.4}{5}\right).31=4,96g\)
\(_{n_P=\dfrac{4,48}{22,4}=0,2=>m_P=0,2.31=6,2}\)
