Đáp án:

\({C_{M{\text{ }}{{\text{H}}_2}S{O_4}}}  = 0,498M\)

\({V_{dd{\text{ }}{\text{KOH}}}}  = 0,996{\text{ lít}}\)

\({m_{dd{\text{ BaC}}{{\text{l}}_2}}}  = 99,6{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({H_2}O + S{O_3}\xrightarrow{{}}{H_2}S{O_4}\)

Ta có:

\({n_{S{O_3}}} = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}}\)

\({m_{dd\;{{\text{H}}_2}S{O_4}{\text{ 65\% }}}} = 200.1,5 = 300{\text{ gam}}\)

\( \to {m_{{H_2}S{O_4}}} = 300.65\%  = 195{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{195}}{{98}} = 1,99{\text{ mol}}\)

\( \to {n_{{H_2}S{O_4}{\text{ trong A}}}} = 1,99 + 0,5 = 2,49{\text{ mol}}\)

\({C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{2,49}}{5} = 0,498M\)

100 ml \(A\) chứa

\({n_{{H_2}S{O_4}}} = 0,1.0,498 = 0,0498{\text{ mol}}\)

Phản ứng xảy ra:

\(2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)

\({n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,0498.2 = 0,0996{\text{ mol}}\)

\({V_{dd{\text{ }}{\text{KOH}}}} = \frac{{0,0996}}{{0,1}} = 0,996{\text{ lít}}\)

50 ml dung dịch \(A\) chứa

\({n_{{H_2}S{O_4}}} = 0,498.0,05 = 0,0249{\text{ mol}}\)

\(BaC{l_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2HCl\)

\( \to {n_{BaC{l_2}}} = {n_{{H_2}S{O_4}}} = 0,0249{\text{ mol}}\)

\( \to {m_{BaC{l_2}}} = 0,0249.(137 + 35,5.2) = 5,1792{\text{ gam}}\)

\( \to {m_{dd{\text{ BaC}}{{\text{l}}_2}}} = \frac{{5,1792}}{{5,2\% }} = 99,6{\text{ gam}}\)

Đáp án:

a)PTHH: SO₃+H₂O→H₂SO₄ (1)

n$SO_{3}$ =$\frac{11.2}{22.4}$ =0.5(mol)

$m_{}$ =200.1,5=300(g)

 (ban đầu)=$\frac{300.65}{100.98}$≈2(mol)

Theo PT (1)

n$H_{2}$$SO_{4}$=$n_{}$ $SO_{3}$ =0.5(mol) 

=>n$H_{2}$$SO_{4}$=0.5+2=2.5(mol)

$C_{M}$= $\frac{2.5}{5}$=0.5(M)

b)PTHH: 2KOH+H₂SO₄→→K₂SO₄+2H₂O (2)

=>n$H_{2}$$SO_{4}$=0,5.0,1=0,05(mol) Theo PT (2)$n_{KOH}$= $2n_{}$ $H_{2}$ $SO_{4}$=0.1(mol) => $V_{KOH}$= $\frac{0.1}{0.1}$ =1(l)

Giải thích các bước giải: