Em tham khảo nha :

\(\begin{array}{l}
a)\\
{K_2}O + {H_2}O \to 2KOH\\
{n_{KOH}} = 0,08 \times 2,5 = 0,2mol\\
{n_{{K_2}O}} = \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{m_{{K_2}O}} = 0,1 \times 94 = 9,4g\\
{m_{CuO}} = 21,4 - 9,4 = 12g\\
b)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuO}} = \dfrac{{12}}{{80}} = 0,15mol\\
{n_{HCl}} = 2{n_{CuO}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{7,3}} = 150g\\
{V_{HCl}} = \dfrac{{150}}{{1,15}} = 130,434ml\\
c)\\
2KOH + C{O_2} \to {K_2}C{O_3}\\
2KOH + S{O_2} \to {K_2}S{O_3}\\
hh:S{O_2}(a\,mol),C{O_2}(b\,mol)\\
\left\{ \begin{array}{l}
2a + 2b = 0,2\\
3a - 2b = 0
\end{array} \right.\\
 \Rightarrow a = 0,04,b = 0,06\\
{m_{hh}} = 0,04 \times 64 + 0,06 \times 44 = 5,2g
\end{array}\)

K₂O + H₂O -> 2KOH (1)

n_KOH=0,2(mol)

Từ 1:

n_K2O=1212n_KOH=0,1(mol)

m_K2O=94.0,1=9,4(g)

m_CuO=21,4-9,4=12(g)

b;

CuO + 2HCl -> CuCl₂ + H₂O (2)

n_CuO=0,15(mol)

Từ 2:

n_Hcl=2n_CuO=0,3(mol)

m_{dd HCl}=0,3.36,57,3%=150(g)0,3.36,57,3%=150(g)

V=150:1,15=130,4(ml)

c;

2KOH + X -> K₂XO + H₂O (3)

Từ 3:

n_X=1212n_KOH=0,1(mol)

n_CO2=0,6(mol)

n_SO2=0,4(mol)

V_CO2=0,6.22,4=13,44(lít)

V_SO2=22,4.0,4=8,96(lít)