Bài giải :
Bài 4 :
`a.`
`-n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{1,2}{56}=0,2(mol)`
Phương trình hóa học :
`Fe+2HCl→FeCl_2+H_2↑`
`0,2` `→` `0,4` `0,2` `0,2` `(mol)`
`⇒V_{H_2}(đktc)=n_{H_2}.22,4=0,2.22,54=4,48(l)`
`b.`
`-m_{HCl}=n_{HCl}.M_{HCl}=0,4.36,5=14,6(g)`
`c.`
`m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,2.127=25,4`(g)`
Bài 5 :
`a`.
`-n_{Al}=\frac{2,4.10^{23}}{6.10^{23}}=0,04(mol)`
$4Al+3O_2\xrightarrow{t^o}2Al_2O_3 $
`0,04→0,03` `0,02` `(mol)`
`⇒V_{O_2}(đktc)=n_{O_2}.22,4=0,03.22,4=0,672(l)`
`⇒V_{kk}=5.V_{O_2}=5.0,672=3,36(l)`
`b.`
`-m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,02.102=2,04(g)`
