8)
\(Ca{(OH)_2} + S{O_2}\xrightarrow{{}}CaS{O_3} + {H_2}O\)
Gọi:
\({n_{S{O_2}}} = x = {n_{CaS{O_3}}} \to {m_{dd{\text{ giảm}}}} = {m_{CaS{O_3}}} - {m_{S{O_2}}} = 120x - 64x = 56x = 5,6 \to x = 0,1 \to V = 0,1.22,4 = 2,24{\text{ lít}}\)
9)
Gọi muối cacbonat có dạng \({R_2}{(C{O_3})_n}\)
\({R_2}{(C{O_3})_n} + 2nHCl\xrightarrow{{}}2RC{l_n} + nC{O_2} + n{H_2}O\)
Ta có:
\({n_{C{O_2}}} = {n_{{H_2}O}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}} \to {n_{HCl}} = 2{n_{C{O_2}}} = 0,6{\text{ mol}} \to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{0,6}}{{0,2}} = 3M\)
BTKL:
\({m_{{R_2}{{(C{O_3})}_n}}} + {m_{HCl}} = {m_{RC{l_n}}} + {m_{C{O_2}}} + {m_{{H_2}O}} \to {m_{RC{l_n}}} = 28,4 + 0,6.36,5 - 0,3.44 - 0,3.18 = 31,7{\text{ gam}}\)
