1)
\(a)Ba + 2{H_2}O\xrightarrow{{}}Ba{(OH)_2} + {H_2}\)
\(b){K_2}O + {H_2}O\xrightarrow{{}}2KOH\)
\(c)F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
\(d)Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
\(e)2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
\(f)4P + 5{O_2}\xrightarrow{{{t^o}}}2{P_2}{O_5}\)
\(g)3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
\(h)C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
\(i){H_2}O + S{O_3}\xrightarrow{{}}{H_2}S{O_4}\)
2)
Phản ứng xảy ra:
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
Ta có:
\({n_{F{e_2}{O_3}}} = \frac{{96}}{{56.2 + 16.3}} = 0,6{\text{ mol}} \to {{\text{n}}_{{H_2}}} = 3{n_{F{e_2}{O_3}}} = 1,8{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 1,8.22,4 = 40,32{\text{ lít}}\)
\({n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,6.2 = 1,2{\text{ mol}} \to {{\text{m}}_{Fe}} = 1,2.56 = 67,2{\text{ gam}}\)
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\({m_{HCl}} = 200.20\% = 40{\text{gam}} \to {{\text{n}}_{HCl}} = \frac{{40}}{{36,5}} = 1,09{\text{ mol < 2}}{{\text{n}}_{Fe}}\)
Vậy Fe dư.
\( \to {n_{Fe{\text{ phản ứng}}}} = {n_{FeC{l_2}}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,545{\text{ mol}}\)
BTKL:
\({m_{dd\;{\text{sau phản ứng}}}} = {m_{Fe}} + {m_{dd{\text{HCl}}}} - {m_{{H_2}}} = 0,545.56 + 200 - 0,545.2 = 229,43{\text{ gam}}\)
\({m_{FeC{l_2}}} = 0,545.(56 + 35,5.2) = 69,215{\text{ gam}} \to {\text{C}}{{\text{\% }}_{FeC{l_2}}} = \frac{{69,215}}{{229,43}} = 30,17\% \)
