Số mol các chất 

`a)` 

$\rm M_{CuSO_4}=64_32+16.4=160(g/mol)$ 

`->`$\rm n_{CuSO_4}=\dfrac{1,6}{160}=0,01(mol)$ 

`b)` 

$\rm M_{K_2SO_4.Al_2(SO_4)_3.24H_2O}=948(g/mol)$ 

`->`$\rm n_{K_2SO_4.Al_2(SO_4)_3.24H_2O}=\dfrac{100}{948}=0,1055(mol)$ 

`c)` 

$\rm n_{SO_3}=\dfrac{22,311}{24,79}=0,9(mol)$ 

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`@` Note: $\rm n=\dfrac{m}{M}$ 

$\rm n(đkc)=\dfrac{V}{24,79}$ 

`a)`

`n_{CuSO_4} = {m}/{M} = {1,6}/{64 + 32 + 16.4} = 0,01 (mol)`

`b)`

`n_{K_2SO_4.Al_2(SO_4)_3}._{24H_2O} = {m}/{M} = {100}/{[(39.2 + 32 + 16.4)+(27.2 + (32 + 16.4).3 + 24.18]`` = {100}/{948} ≈ 0,105 (mol)`

`c)`

`V_{SO_3} = {V}/{24,79} = {22,311}/{24,79} = 0,9 (mol)`