Làm hộ mk 2 bài này gấp vs ak !!

3)

a)

\(4Na + {O_2}\xrightarrow{{{t^o}}}2N{a_2}O\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)

\(N{a_2}C{O_3} + 2HCl\xrightarrow{{}}2NaCl + C{O_2} + {H_2}O\)

\(2NaCl + 2{H_2}O\xrightarrow{{dpdd/cmn}}2NaOH + C{l_2} + {H_2}\)

\(NaOH + C{O_2}\xrightarrow{{}}NaHC{O_3}\)

b)

\(2Cu + {O_2}\xrightarrow{{{t^o}}}2CuO\)

\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)

\(CuS{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + CuC{l_2}\)

\(CuC{l_2} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + 2NaCl\)

\(Cu{(OH)_2} + 2HN{O_3}\xrightarrow{{}}Cu{(N{O_3})_2} + 2{H_2}O\)

\(Cu{(N{O_3})_2} + Fe\xrightarrow{{}}Fe{(N{O_3})_2} + Cu\)

c)

\(4Na + {O_2}\xrightarrow{{{t^o}}}2N{a_2}O\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)

\(N{a_2}C{O_3} + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + C{O_2} + {H_2}O\)

\(N{a_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2NaCl\)

4)

Phản ứng xảy ra:

\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)

Ta có:

\({n_{CuO}} = \frac{8}{{64 + 16}} = 0,1{\text{ mol}}\)

\({m_{{H_2}S{O_4}}} = 100.19,6\%  = 19,6{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{19,6}}{{98}} = 0,2{\text{ mol > }}{{\text{n}}_{CuO}}\)

Vậy \(H_2SO_4\) dư.

\( \to {n_{{H_2}S{O_4}{\text{ dư}}}} = 0,2 - 0,1 = 0,1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}{\text{ dư}}}} = 0,1.98= 9,8{\text{ gam}}\)

\({n_{CuS{O_4}}} = {n_{CuO}} = 0,1{\text{ mol}}\)

\( \to {m_{CuS{O_4}}} = 0,1.(64 + 96) = 16{\text{ gam}}\)

BTKL:

\({m_{CuO}} + {m_{dd{{\text{H}}_2}S{O_4}}} = {m_{dd}} = 100 + 8 = 108{\text{ gam}}\)

\( \to C{\% _{CuS{O_4}}} = \frac{{16}}{{108}} = 14,815\% \)

\(C{\% _{{H_2}S{O_4}{\text{ dư}}}} = \frac{{9,8}}{{108}} = 9,07\% \)