Đáp án:

\( {m_{CaO}} ) = 112{\text{ gam}}\)

\({m_{{H_2}O}} = 36{\text{ gam}}\)

\({m_{Mn{O_2}}}  = 174{\text{ gam}}\)

\({V_{dd\;{{\text{H}}_2}S{O_4}}}  = 696,65{\text{ c}}{{\text{m}}^3}\)

Giải thích các bước giải:

Cần hết cả mấy chất trên luôn

Các phản ứng xảy ra để điều chế:

\(CaO + {H_2}O\xrightarrow{{}}Ca{(OH)_2}\)

\(NaCl + {H_2}S{O_4}\xrightarrow{{{t^o}}}NaHS{O_4} + HCl\)

\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)

\(Ca{(OH)_2} + C{l_2}\xrightarrow{{}}CaOC{l_2} + {H_2}O\)

Clorua vôi là \(CaOCl_2\)

\({n_{CaOC{l_2}}} = \frac{{254}}{{40 + 16 + 35,5.2}} = 2{\text{ mol}}\)

Theo phản ứng:

\({n_{CaOC{l_2}}} = {n_{Ca{{(OH)}_2}}} = {n_{C{l_2}}}\)

\({n_{Ca{{(OH)}_2}}} = {n_{CaO}} = {n_{{H_2}O}} = 2{\text{ mol}}\)

\( \to {m_{CaO}} = 2.(40 + 16) = 112{\text{ gam}}\)

\({m_{{H_2}O}} = 2.18 = 36{\text{ gam}}\)

\({n_{C{l_2}}} = {n_{Mn{O_2}}} = 2{\text{ mol;}}{{\text{n}}_{HCl}} = 4{n_{C{l_2}}} = 8{\text{ mol}}\)

\( \to {m_{Mn{O_2}}} = 2.(55 + 16.2) = 174{\text{ gam}}\)

\({n_{NaCl}} = {n_{{H_2}S{O_4}}} = {n_{HCl}} = 8{\text{ mol}}\)

\( \to {m_{NaCl}} = 8.58,5 = 468{\text{ gam}}\)

\({m_{{H_2}S{O_4}}} = 8.98 = 784{\text{ gam}}\)

\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{784}}{{70\% }} = 1120{\text{ gam}}\)

\( \to {V_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{1120}}{{1,61}} = 696,65{\text{ c}}{{\text{m}}^3}\)