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`-` `Fe + 2HCl -> FeCl_2 + H_2 ↑`

a. `-` `m_{HCl} = \frac{200 xx 10}{100} = 20` `(g)`

`-` `n_{HCl} = \frac{20}{36,5} = \frac{40}{73}` `(mol)`

`-` Theo phương trình `n_{Fe} = \frac{20}{73}` `(mol)`

`->` `m_{Fe} = \frac{20}{73} xx 56 = 15,34` `(g)`

b. `-` Theo phương trình `n_{FeCl_2} = \frac{20}{73}` `(mol)`

`->` `m_{FeCl_2} = \frac{20}{73} xx 127 = 34,75` `(g)`

`-` Bảo toàn khối lượng, ta có:

`-` `m_{Fe} + m_{dd} = m_{ddspu} + m_{H_2}`

`->` `m_{ddspu} = 15,34 + 200 - (\frac{20}{73} xx 2) = 214,79` `(g)`

`->` `C%_{FeCl_2} = \frac{34,75}{214,79} xx 100% = 16,17%`

Đáp án:

$a,m_{Fe}=15,4\,(g)$ 

$b,C\%=16,2\%$

Giải thích các bước giải:

$Fe+2HCl\rightarrow FeCl_2+H_2$

$a,m_{HCl}=\dfrac{200.10}{100}=20\,(g)$

$⇒n_{HCl}=\dfrac{20}{36,5}=\dfrac{40}{73}\,(mol)$

$⇒n_{Fe}=\dfrac{1}{2}.n_{HCl}=\dfrac{20}{73}\,(mol)$

$⇒m_{Fe}=\dfrac{20}{73}.56=\dfrac{1120}{73}=15,4\,(g)$

$b,n_{H_2}=n_{Fe}=\dfrac{20}{73}\,(mol)$

$⇒m_{H_2}=\dfrac{20}{73}.2=\dfrac{40}{73}\,(g)$

$⇒m_{dd}=\dfrac{1120}{73}+200-\dfrac{40}{73}=\dfrac{15680}{73}=214,8\,(g)$

$n_{FeCl_2}=n_{Fe}=\dfrac{20}{73}\,(mol)$

$⇒m_{FeCl_2}=\dfrac{20}{73}.127=\dfrac{2540}{73}=34,8\,(g)$

$⇒C\%=\dfrac{34,8.100\%}{214,8}=16,2\%$.