Đáp án:
a=3,571 gam
\({\text{C}}{{\text{\% }}_{{K_2}S{O_4}}} = 3,846\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KOH + S{O_3}\xrightarrow{{}}{K_2}S{O_4} + {H_2}O\)
Ta có:
\({m_{KOH}} = 100.5\% = 5{\text{ gam}} \to {{\text{n}}_{KOH}} = \frac{5}{{56}} \to {n_{S{O_3}}} = {n_{{K_2}S{O_4}}} = \frac{1}{2}{n_{KOH}} = \frac{5}{{112}} \to a = \frac{5}{{112}}.80 = 3,571{\text{ gam}}\)
\({m_{{K_2}S{O_4}}} = \frac{5}{{112}}.(39.2 + 96) = 7,768{\text{ gam}} \to {\text{C}}{{\text{\% }}_{{K_2}S{O_4}}} = \frac{{7,768}}{{202}} = 3,846\% \)