https://i.imgur.com/Lz15304.jpg
2Pb(NO3)2 | → | 4NO2 | + | 4O2 | + | 2PbO |
Ta có
n Pb(NO3)2=66,2/331=0,2(mol)
Theo pthh
n PhO=n Pb(NO3)2=0,2(mol)
m PbO=0,2.223=44,6(g)
H=\(\frac{44,6}{55,4}.100\%=80,5\%\)