Đáp án:

 1. \( \text C\%_{HCl}=2,42\%; \text C\%_{H_2SO_4\ \text{dư}}=2,57\%\)

Giải thích các bước giải:

 Bài 1.

\(\begin{array}{l}m_{H_2SO_4}=180\cdot 15\%=27\ \text{gam}⇒ n_{H_2SO_4}=\frac{27}{98}\ \text{mol}\\ m_{BaCl_2}=320\cdot 10\%=32\ \text{gam}⇒ n_{BaCl_2}=\frac{32}{208}=\frac 2{13}\ \text{mol}\\ H_2SO_4+BaCl_2\to BaSO_4+2HCl\end{array}\)

Vì \(\dfrac{2}{13}<\dfrac{27}{98}\) nên sau phản ứng \(BaCl_2\) hết

\(⇒ n_{HCl}=2n_{BaCl_2}=\frac{2}{13}\cdot 2=\frac{4}{13} \ \text{mol}; n_{H_2SO_4\ \text{phản ứng}}=n_{BaCl_2}=\frac{2}{13}\ \text{mol}⇒ n_{H_2SO_4\ \text{dư}}=\frac{27}{98}-\frac 2{13}=\frac{155}{1274}\ \text{mol}; n_{\text{Ba}SO_4}=n_{BaCl_2}=\frac 2{13}\ \text{mol}\)

\(⇒ m_{HCl}=36,5\cdot \frac 4{13}=\frac{146}{13}\ \text{gam}; m_{H_2SO_4\ \text{dư}}=\dfrac{155}{1274}\cdot 98=\dfrac{155}{13}\ \text{gam}; m_{BaSO_4}=\frac 2{13}\cdot 233=\frac{466}{13}\ \text{gam}\)

BTKL: \(m_{\text{dung dịch}\ H_2SO_4}+m_{\text{dung dịch}\ BaCl_2}=m_{\text{dung dịch sau phản ứng}}+m_{BaSO_4}\)

\(⇒ m_{\text{dung dịch sau phản ứng}}=180+320-\frac{466}{13}=\frac{6034}{13}\ \text{gam}\)

\(⇒ \text C\%_{HCl}=\dfrac{\frac{146}{13}}{\frac{6034}{13}}\cdot 100\%=2,42\%; \text C\%_{H_2SO_4\ \text{dư}}=\cfrac{\frac{155}{13}}{\frac{6034}{13}}\cdot 100\%=2,57\%\)

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{m_{{H_2}S{O_4}}} = \dfrac{{180 \times 15\% }}{{100\% }} = 27g\\
{m_{BaC{l_2}}} = \dfrac{{320 \times 10\% }}{{100\% }} = 32g\\
 \to {n_{{H_2}S{O_4}}} = 0,28mol\\
 \to {n_{BaC{l_2}}} = 0,15mol\\
 \to {n_{{H_2}S{O_4}}} > {n_{BaC{l_2}}} \to {n_{{H_2}S{O_4}}}dư\\
 \to {n_{HCl}} = 2{n_{BaC{l_2}}} = 0,3mol\\
 \to {m_{HCl}} = 10,95g\\
 \to {n_{{H_2}S{O_4}}}(pt) = {n_{BaC{l_2}}} = 0,15mol\\
 \to {n_{{H_2}S{O_4}}}dư = 0,13mol\\
 \to {m_{{H_2}S{O_4}}}dư = 12,74g\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,15mol\\
 \to {m_{BaS{O_4}}} = 34,95g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}{H_2}S{O_4}}} + {m_{{\rm{dd}}BaC{l_2}}} - {m_{BaS{O_4}}} = 180 + 320 - 34,95 = 465,05g\\
 \to C{\% _{{H_2}S{O_4}}}dư = \dfrac{{12,74}}{{465,05}} \times 100\%  = 2,74\% \\
 \to C{\% _{HCl}} = \dfrac{{10,95}}{{465,05}} \times 100\%  = 2,35\% 
\end{array}\)

\(\begin{array}{l}
2.\\
MgC{l_2} + Ba{(OH)_2} \to Mg{(OH)_2} + BaC{l_2}\\
Mg{(OH)_2} \to MgO + {H_2}O\\
a)\\
{n_{MgC{l_2}}} = 0,2mol\\
{n_{Ba{{(OH)}_2}}} = 0,225mol\\
 \to {n_{Ba{{(OH)}_2}}} > {n_{MgC{l_2}}} \to {n_{Ba{{(OH)}_2}}}du\\
 \to {n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = 0,2mol\\
 \to {n_{MgO}} = {n_{Mg{{(OH)}_2}}} = 0,2mol\\
 \to {m_{MgO}} = 8g\\
b)\\
{n_{BaC{l_2}}} = {n_{MgC{l_2}}} = 0,2mol\\
 \to {m_{BaC{l_2}}} = 41,6g\\
{n_{Ba{{(OH)}_2}}}(pt) = {n_{MgC{l_2}}} = 0,2mol\\
 \to {n_{Ba{{(OH)}_2}}}(dư) = 0,025mol\\
 \to {m_{Ba{{(OH)}_2}}}(dư) = 4,275g\\
 \to {m_{{\rm{dd}}Ba{{(OH)}_2}}} = 150 \times 1,12 = 168g\\
 \to {m_{{\rm{ddthuđược}}}} = {m_{MgC{l_2}}} + {m_{{\rm{dd}}Ba{{(OH)}_2}}} - {m_{Mg{{(OH)}_2}}} = 175,4g\\
b)\\
C{M_{BaC{l_2}}} = \dfrac{{0,2}}{{0,1 + 0,15}} = 0,8M\\
C{M_{Ba{{(OH)}_2}}}dư = \dfrac{{0,025}}{{0,1 + 0,15}} = 0,1M\\
C{\% _{BaC{l_2}}} = \dfrac{{41,6}}{{175,4}} \times 100\%  = 23,7\% \\
C{\% _{Ba{{(OH)}_2}}}dư = \dfrac{{4,275}}{{175,4}} \times 100\%  = 2,44\% 
\end{array}\)