Em tham khảo nha:

\(\begin{array}{l}
1)\\
nCuO = \dfrac{8}{{80}} = 0,1\,mol\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
nHCl = 2nCuO = 0,2\,mol\\
{C_M}HCl = \dfrac{{0,2}}{{0,5}} = 0,4M\\
2)\\
nC{O_2} = nCaC{O_3} = \dfrac{{20}}{{100}} = 0,2\,mol\\
\% VC{O_2} = \dfrac{{0,2 \times 22,4}}{{0,2 \times 22,4 + 4,48}} \times 100\%  = 50\% \\
3)\\
nNaOH = 0,5 \times 2 = 1\,mol\\
nHCl = 0,4 \times 5 = 2\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
\dfrac{1}{1} < \dfrac{2}{1} \Rightarrow \text{ HCl dư nên quỳ tím hóa đỏ}\\
4)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
nC{l_2} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
nFeC{l_3} = \dfrac{{0,3 \times 2}}{3} = 0,2\,mol\\
mFeC{l_3} = 0,2 \times 162,5 = 32,5g
\end{array}\)