Question

Đốt cháy 6,2g P thu được a(g) điphotphopenta oxit(P2O5) . Tính a

Answer 1

4P + 5O₂ \(\rightarrow\)2P₂O₅

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

Theo PTHH ta có:

\(\dfrac{1}{2}\)n_P=n_P2O5=0,1(mol)

m_P2O5=142.0,1=14,2(g)

Answer 2

PTHH: 4P+5O₂---->2P₂O₅

n_P= 6,2/31 =0,2 (mol)

theo pt: n_P2O5= \(\dfrac{1}{2}\). n_P=\(\dfrac{1}{2}\).0,2=0,1 (mol)

=> m_P2O5= 0,1.142=14,2 (g)