Giải thích các bước giải: a,`BaCl_2`+`H_2SO_4`→`BaSO_4`+`2HCl`

                                                                     

Ta có `m_(BaCl_2)`=400.5,2%=20,8(g)

⇒`n(BaCl_2)`=$\frac{20,8}{208}$ =0,1(mol)

Ta có `m_dd(H_2SO_4)` =1,14.100=114(g)

⇒`m_(H_2SO_4)` =114.20%=22,8

⇒`n_(H_2SO_4)` =22,8/98≈0,233(mol)

Theo tỷ lệ ta thấy `BaCl_2` hết, `H_2SO_4` dư

Theo PTHH `n_(BaSO_4)` =0,1(mol)

⇒m↓=0,1.233=23,3(g)

b,ta có `m_dd`=400+114-23,3=490,7(g)

`m_(H_2SO_4) phản ứng=0,1.98=9,8(g)

⇒`m_(H_2SO_4)` dư=22,8-9,8=13(g)

⇒C%`H_2SO_4`=$\frac{13}{490,7}$ .100%≈2,65%

Chúc bạn học tốt.

 

Giải thích các bước giải:

\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to B{\rm{aS}}{{\rm{O}}_4} + 2HCl\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 5,2\% }}{{100\% }} = 20,8g \to {n_{BaC{l_2}}} = 0,1mol\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = 100 \times 1,14 = 114g\\
 \to {m_{{H_2}S{O_4}}} = \dfrac{{114 \times 20\% }}{{100\% }} = 22,8g \to {n_{{H_2}S{O_4}}} = 0,23mol\\
 \to {n_{{H_2}S{O_4}}} > {n_{BaC{l_2}}} \to {n_{{H_2}S{O_4}}}dư\\
a)\\
{n_{{\rm{BaS}}{O_4}}} = {n_{BaC{l_2}}} = 0,1mol \to {m_{{\rm{BaS}}{O_4}}} = 23,3g\\
b)\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol \to {m_{HCl}} = 7,3g\\
 \to {n_{{H_2}S{O_4}(pt)}} = {n_{BaC{l_2}}} = 0,1mol\\
 \to {n_{{H_2}S{O_4}}}dư = 0,13mol \to {m_{{H_2}S{O_4}}}du = 12,74g\\
{m_{{\rm{dd}}thuđược}} = {m_{{\rm{dd}}BaC{l_2}}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{\rm{BaS}}{O_4}}} = 490,7g\\
 \to C{\% _{HCl}} = \dfrac{{7,3}}{{490,7}} \times 100\%  = 1,5\% \\
 \to C{\% _{{H_2}S{O_4}}}dư = \dfrac{{12,74}}{{490,7}} \times 100\%  = 2,6\% 
\end{array}\)