Bài 1:
a) PTHH: 2Na + 2H2O --> 2NaOH + H2
b) Theo ĐLBTKL: \(m_{Na}+m_{H_2O}=m_{NaOH}+m_{H_2}\)
=> mH2 = 9,2 + 7,2 - 16 = 0,4 (g)
c) \(n_{H_2}=\dfrac{0,4}{2}=0,2\left(mol\right)=>V_{H_2}=0,2.22,4=4,48\left(l\right)\)
Bài 2:
\(m_{Fe\left(pư\right)}=8,4-2,8=5,6\left(g\right)\)
Theo ĐLBTKL: \(m_{Fe\left(pư\right)}+m_{S\left(pư\right)}=m_{FeS}\)
=> a = 5,6 + 3,2 = 8,8 (g)
Bài 3:
a) \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b) Ba + 2H2O --> Ba(OH)2 + H2\(\uparrow\)
c) \(3C+2Fe_2O_3\underrightarrow{t^o}4Fe+3CO_2\)
d) \(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow3Na_2SO_4+2Fe\left(OH\right)_3\downarrow\)
e) \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
f) \(K_3PO_4+3AgNO_3\rightarrow Ag_3PO_4+3KNO_3\)
g) \(3Ba\left(NO_3\right)_2+Al_2\left(SO_4\right)_3\rightarrow3BaSO_4\downarrow+2Al\left(NO_3\right)_3\)
h) \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)
i) \(4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\)
