Giúp mình vs

`-` Hướng làm $1:$

$n_{CO_2}=n_{BaCO_3}=\dfrac{27,58}{197}=0,14(mol)$

$n_{Y}=0,315(mol)$

`=>`$n_{H_2}=0,315-0,14=0,175(mol)$

$n_{MgCl_2}=a(mol)$

$n_{NaCl}=2,55a(mol)$

`=>`$n_{HCl}=n_{Cl^-}=4,55a(mol)$

`-` Bảo toàn nguyên tố $H:$

$n_{H_2O}=\dfrac{n_{HCl}-2n_{H_2}}{2}=\dfrac{4,55a-0,35}{2}$

`-` Bảo toàn khối lượng:

$m+4,55a \times 36,5=m+21,665+0,175 \times 2+0,14 \times 44+18 \times \dfrac{4,55a-0,35}{2}$

`=>`$a=0,2$

`=>`$m=2,55 \times 0,2 \times  23 + 0,2 \times 24 +35,5 \times 4,55 \times 0,2$

$=27,17(gam)$

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`-` Hướng làm $2:$

$C_2:$

$n_{MgCl_2}=a(mol)$

$n_{NaCl}=2,55a(mol)$

`=>`$n_{Cl^-}=4,55a(mol)$

$n_{CO_2}=n_{BaCO_3}=\dfrac{27,58}{197}=0,14(mol)$

$n_{Y}=0,315(mol)$

`=>`$n_{H_2}=0,315-0,14=0,175(mol)$

`=>`$n_{O,bơm}=\dfrac{0,175 \times 2 }{2}=0,175(mol)$

`-` Bơm thêm $0,0875(mol) \,\, O:$

`->`$m+2,8\begin{cases} Na:2,55a\\Mg:a\\CO_3:0,14\\O:2,275a-0,14 \end{cases}$

`-` Muối: 

$m+21,665 \begin{cases} Na:2,55a\\Mg:a\\Cl:4,55a \end{cases}$

`=>`$60 \times 0,14+16 \times (2,275a-0,14)-35,5a=m+2,8-m-21,665$

`=>`$a=0,2$

`=>`$m=2,55 \times 0,2 \times  23 + 0,2 \times 24 +35,5 \times 4,55 \times 0,2$

$=27,17(gam)$

Đáp án: $m=27,17g$

Giải thích các bước giải:

$n_Y=\dfrac{7,056}{22,4}=0,315$ mol 

$n_{BaCO_3}=\dfrac{27,58}{197}=0,14$ mol 

$\Rightarrow n_{CO_2}= n_{BaCO_3}=0,14$ mol 

$\Rightarrow n_{H_2}=0,315-0,14=0,175$ mol 

Đặt $n_{MgCl_2}=x$ mol 

$\Rightarrow n_{NaCl}=2,55x$ mol  

$\Rightarrow n_{HCl}=n_{NaCl}+2n_{MgCl_2}=2,55x+2x=4,55x$ mol 

Bảo toàn $H$:

$n_{HCl}=2n_{H_2}+2n_{H_2O}$ 

$\Rightarrow n_{H_2O}=\dfrac{4,55x}{2}-0,175=2,275x-0,175$ mol 

Bảo toàn khối lượng:

$m+36,5.4,55x= m+21,665+0,14.44+0,175.2+18(2,275x-0,175)$ 

$\Rightarrow x=0,2$ 

$n_{Mg}=0,2$ mol 

$n_{Na}=0,2.2,55=0,51$ mol 

$n_{CO_3}=n_{CO_2}=0,14$ mol 

Bảo toàn $O$:

$3n_{CO_3}+ n_O=2n_{CO_2}+ n_{H_2O}$ 

$\Rightarrow n_O=2.0,14+2,275.0,2-0,175-3.0,14=0,14$ mol 

Vậy $m=0,2.24+0,51.23+0,14.60+0,14.16=27,17g$