Question

đốt cháy P trong 4,48 (lít) oxi đktc tìm khối lượng P đã dùng (P31)

 

Answer 1

\(n_{O2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Pt : \(4P+5O_2\underrightarrow{t^o}2P_2O_5|\)

        4        5           2

       0,16   0,2

\(n_P=\dfrac{0,2.4}{5}=0,16\left(mol\right)\)

⇒ \(m_P=0,16.31=4,96\left(g\right)\)

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Answer 2

\(4P+5O_2\rightarrow2P_2O_5\)

\(n_{O_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(\Leftrightarrow n_{P_2O_5}=0.5\left(mol\right)\)

\(\Leftrightarrow n_P=1\left(mol\right)\)

\(m_P=1\cdot31=31\left(g\right)\)

 

Answer 3

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

Answer 4

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)

\(m_P=n.M=\left(\dfrac{0,2.4}{5}\right).31=4,96g\)

Answer 5

\(_{n_P=\dfrac{4,48}{22,4}=0,2=>m_P=0,2.31=6,2}\)