Đáp án:
\({m_{rắn}} = 59,5{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{C{O_2}}} = \frac{{7,84}}{{22,4}} = 0,35{\text{ mol}}\)
\({m_{KOH}} = 210.24\% = 50,4{\text{ gam}}\)
\( \to {n_{KOH}} = \frac{{50,4}}{{56}} = 0,9{\text{ mol}}\)
\( \to \frac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,9}}{{0,35}} > 2\)
Vậy \(KOH\) dư.
\(2KOH + C{O_2}\xrightarrow{{}}{K_2}C{O_3} + {H_2}O\)
Ta có:
\({n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,35{\text{ mol}}\)
\( \to {n_{KOH{\text{ dư}}}} = 0,9 - 0,35.2 = 0,2{\text{ mol}}\)
\( \to {m_{rắn}} = {m_{{K_2}C{O_3}}} + {m_{KOH}}\)
\( = 0,35.(39.2 + 60) + 0,2.56 = 59,5{\text{ gam}}\)
