Đáp án:
\({{\text{V}}_{S{O_2}}} = 4,48{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(N{a_2}S{O_3} + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + S{O_2} + {H_2}O\)
Ta có:
\({n_{N{a_2}S{O_3}}} = \frac{{25,2}}{{23.2 + 32 + 16.3}} = 0,2{\text{ mol;}}{{\text{m}}_{{H_2}S{O_4}}} = 19,6\% .200 = 39,2{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{39,2}}{{98}} = 0,4{\text{ mol > }}{{\text{n}}_{N{a_2}S{O_3}}}\)
\( \to {n_{S{O_2}}} = {n_{N{a_2}S{O_3}}} = 0,2{\text{ mol}} \to {{\text{V}}_{S{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
BTKL:
\({m_{dd}} = {m_{N{a_2}S{O_3}}} + {m_{dd{\text{ }}{H_2}S{O_4}}} - {m_{S{O_2}}} = 25,2 + 200 - 0,2.64 = 212,4{\text{ gam}}\)
\({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2{\text{ mol}} \to {{\text{m}}_{N{a_2}S{O_4}}} = 0,2.(23.2 + 96) = 28,4{\text{ gam;}}{{\text{m}}_{{H_2}S{O_4}}} = 0,2.98 = 19,6{\text{ gam}}\)
\( \to C{\% _{N{a_2}S{O_4}}} = \frac{{28,4}}{{212,4}} = 13,37\% ;C{\% _{{H_2}S{O_4}}} = \frac{{19,6}}{{212,4}} = 9,23\% \)
