Em tham khảo nha :
\(\begin{array}{l}
1)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{NaOH}} = 2{n_{C{O_2}}} = 0,6mol\\
{V_{NaOH}} = \dfrac{{0,6}}{2} = 0,3l = 300ml\\
2)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
hh:CuO(a\,mol),F{e_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
80a + 160b = 3,2\\
a - 2b = 0
\end{array} \right.\\
\Rightarrow a = 0,02;b = 0,01\\
{m_{CuO}} = 0,02 \times 80 = 1,6g\\
{m_{F{e_2}{O_3}}} = 3,2 - 1,6 = 1,6g
\end{array}\)