Em tham khảo nha:
\(\begin{array}{l}
3)\\
CTHH:{M_x}{O_y}\\
\% {m_O} = 40\% \Leftrightarrow \dfrac{{16y}}{{M \times x + 16y}} = 40\% \\
\Leftrightarrow 0,4Mx + 6,4y = 16y \Rightarrow M = \dfrac{{12 \times 2y}}{x}(g/mol)\\
\dfrac{{2y}}{x} = 2 \Rightarrow M = 12 \times 2 = 24g/mol \Rightarrow M:Magie(Mg)\\
\dfrac{{2y}}{x} = 2 \Rightarrow x:y = 2:2 = 1:1 \Rightarrow CTHH:MgO\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{MgO}} = \dfrac{{10}}{{40}} = 0,25\,mol\\
{n_{HCl}} = 2{n_{MgO}} = 0,5\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,5 \times 36,5}}{{7,3\% }} = 250g\\
4)\\
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
b)\\
{n_{{H_2}S{O_4}}} = 0,1 \times 3,5 = 0,35\,mol\\
hh:CuO(a\,mol),F{e_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
80a + 160b = 20\\
a + 3b = 0,35
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,1\\
{m_{CuO}} = 0,05 \times 80 = 4g\\
{m_{F{e_2}{O_3}}} = 20 - 4 = 16g\\
5)\\
a)\\
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Ba{{(OH)}_2}}} = {n_{C{O_2}}} = 0,2\,mol\\
{C_M}Ba{(OH)_2} = \frac{{0,2}}{{0,5}} = 0,4M\\
c)\\
{n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,2\,mol\\
{m_{BaC{O_3}}} = 0,2 \times 197 = 39,4g
\end{array}\)
