Question

Answer 1

a.\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PTHH : 4P + 5O₂ \(\underrightarrow{t^0}\)2P₂O₅

            0,2      0,25

\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

Answer 2

4P+5O₂-to>2P₂O₅

0,2----0,25 mol

n _P=\(\dfrac{6,2}{31}\)=0,2 mol

=>V_O2=0,25.22,4=5,6l

 

 

Answer 3