12)

 Phản ứng xảy ra:

\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)

Ta có:

\({n_{CaC{O_3}}} = \frac{{15}}{{100}} = 0,15{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)

\( \to a = {V_{C{O_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

Ta có:

\({n_{HCl}} = 2{n_{C{O_2}}} = 0,15.2 = 0,3{\text{ mol}}\)

\( \to {V_{dd{\text{HCl}}}} = \frac{{0,3}}{1} = 0,3{\text{ lit}}\)

Dẫn \(CO_2\) vào nước vôi trong.

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

Ta có:

\({n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15{\text{ mol}}\)

\( \to {m_{CaC{O_3}}} = 0,15.100 = 15{\text{ gam}}\)

13)

Câu này để cho thể tích dung dịch axit rồi thì hỏi gì nữa em ?

Phản ứng xảy ra:

\(Ba{(OH)_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2{H_2}O\)

Ta có:

\({m_{BaC{l_2}}} = 400.5,2\%  = 20,8{\text{ gam}}\)

\( \to {n_{BaC{l_2}}} = \frac{{20,8}}{{137 + 35,5.2}} = 0,1{\text{ mol}}\)

Ta có:

\({m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = 100.1,14 = 114{\text{ gam}}\)

\( \to {m_{{H_2}S{O_4}}} = 114.20\%  = 22,8{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{22,8}}{{98}} = 0,23265{\text{ mol > }}{{\text{n}}_{BaC{l_2}}}\)

\( \to {n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1{\text{ mol}}\)

\(\to {m_{BaS{O_4}}} = 0,1.(137 + 96) = 23,3{\text{ gam}}\)