`flower`

Đáp án + Giải thích các bước giải:

`1.`

 `a.`

`n_{Zn}=13/65=0,2(mol)`

`n_{Al}=(5,4)/27=0,2(mol)`

`n_{Mg}=(4,8)/240,2(mol)`

`n_{H_2SO_4}=(4,9)/98=0,05(mol)`

`n_{NaOH}=20/40=0,5(mol)`

`n_{Al_2O_3}=51/102=0,5(mol)`

`n_{Fe(OH)_2}=45/90=0,5(mol)`

`n_{Al_2(SO_4)_3}=171/342=0,5(mol)`

`*` CT tính số mol `(1)` : `n=m/M`

`-` `n` : số mol `(mol)`

`-` `m` : khối lượng `(g)`

`-` `M` : khối lượng mol `(g//mol)`

 `b.`

`n_{O_2}=(2,24)/(22,4)=0,1(mol)`

`n_{CO_2}=(3,36)/(22,4)=0,15(mol)`

`n_{SO_2}=(67,2)/(22,4)=3(mol)`

`n_{NO_2}=(5,6)/(22,4)=0,25(mol)`

`n_{NH_3}=(13,44)/(22,4)=0,6(mol)`

`n_{CH_4}=(8,96)/(22,4)=0,4(mol)`

`*` CT tính số mol ở điều kiện tiêu chuẩn `(2)` : `n=V/(22,4)`

`-` `n` : số mol `(mol)`

`-` `V` : thể tích `(l)`

$a)n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2(mol)$

$n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2(mol)$

$n_{Mg}=\dfrac{m}{M}=\dfrac{4,8}{24}=0,2(mol)$

$n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{4,9}{98}=0,05(mol)$

$n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5(mol)$

$n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{51}{102}=0,5(mol)$

$n_{Fe(OH)_2}=\dfrac{m}{M}=\dfrac{45}{90}=0,5(mol)$

$n_{Al_2(SO_4)_3}=\dfrac{m}{M}=\dfrac{171}{342}=0,5(mol)$

$b)n_{O_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1(mol)$

$n_{CO_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15(mol)$

$n_{SO_2}=\dfrac{V}{22,4}=\dfrac{67,2}{22,4}=3(mol)$

$n_{NO_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25(mol)$

$n_{NH_3}=\dfrac{V}{22,4}=\dfrac{13,44}{22,4}=0,6(mol)$

$n_{CH_4}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4(mol)$