1)
Ta có:
\({n_{C{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol;}}{{\text{n}}_{KOH}} = 0,1.1 = 0,1{\text{ mol}}\)
\( \to \frac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,1}}{{0,1}} = 1\)
Vậy phản ứng tạo muối \(KHCO_3\)
\(KOH + C{O_2}\xrightarrow{{}}KHC{O_3}\)
Ta có:
\({n_{KHC{O_3}}} = {n_{KOH}} = 0,1{\text{ mol}}\)
\( \to {m_{KHC{O_3}}} = 0,1.(39 + 1 + 60) = 10{\text{ gam}}\)
2)
Ta có:
\({n_{C{O_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)
\({n_{NaOH}} = 0,2.2,5 = 0,5{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,5}}{{0,3}} = 1,67\)
Do vậy phản ứng tạo 2 muối.
\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)
\(NaOH + C{O_2}\xrightarrow{{}}NaHC{O_3}\)
Gọi số mol \(Na_2CO_3;NaHCO_3\) lần lượt là \(x;y\)
\( \to {n_{C{O_2}}} = x + y = 0,3{\text{ mol}}\)
\({n_{NaOH}} = 2x + y = 0,5\)
Giải được: \(x=0,2;y=0,1\)
\( \to m = {m_{N{a_2}C{O_3}}} + {m_{NaHC{O_3}}} = 0,2.(23.2 + 60) + 0,1.(23 + 1 + 60) = 29,6{\text{ gam}}\)
3)
Ta có:
\({n_{C{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
\({n_{KOH}} = 0,2.1 = 0,2{\text{ mol}}\)
\( \to \frac{{{n_{KOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,2}}{{0,1}} = 2\)
Vậy phản ứng tạo muối \(K_2CO_3\)
\(2KOH + C{O_2}\xrightarrow{{}}{K_2}C{O_3} + {H_2}O\)
Ta có:
\({n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,1{\text{ mol}}\)
\( \to {m_{{K_2}C{O_3}}} = 0,1.(39.2 + 60) = 13,8{\text{ gam}}\)
