a,

$n_{Fe_2(SO_4)_3}=\dfrac{3.10^{23}}{6.10^{23}}=0,5(mol)$

$\Rightarrow m_{Fe_2(SO_4)_3}=0,5.400=200g$

b,

Số phân tử $Al(NO_3)_3=0,25.6.10^{23}=1,5.10^{23}$

$\Rightarrow n_{Al(NO_3)_3}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25 (mol)$

$\Rightarrow m_{Al(NO_3)_3}=0,25.213=53,25g$

c,

$n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15(mol)$

$\Rightarrow m_{C_2H_4}=0,15.28=4,2g$

`a,`

`n_(Fe_2(SO_4)_3)=\frac{3.10^23}{6.10^23}=0,5(mol)`

`=>m_(Fe_2(SO_4)_3)=0,5.400=200(g)`

`b,`

`n_(Al(NO_3)_3)=\frac{0,25.6.10^23}{6.10^23}=0,25(mol)`

`=>m_(Al(NO_3)_3)=0,25.213=53,35(g)`

`c,`

`n_(C_2H_4)=\frac{3,36}{22,4}=0,15(mol)`

`m_(C_2H_4)=0,15.28=4,2(g)`