Em tham khảo nha :

\(\begin{array}{l}
6)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{Cu}} = {n_{{H_2}}} = 0,2mol\\
{m_{Cu}} = 0,2 \times 64 = 12,8g\\
8)\\
N{a_2}O + {H_2}O \to 2NaOH\\
P1:\\
C{O_2} + NaOH \to NaHC{O_3}\\
P2:\\
NaOH + NaHC{O_3} \to N{a_2}C{O_3} + {H_2}O\\
N{a_2}C{O_3} + Ca{(N{O_3})_2} \to CaC{O_3} + 2NaN{O_3}
\end{array}\)

6.

nFe=11,2/56=0,2mol

Fe+2HCl→FeCl2+H2

VH2=0,2.22,4=4,48l

H2+CuO→Cu+H2O

mCu=0,2.64=12,8g

8.

Na2O+H2O→2NaOH

P1+CO2 dư

NaOH+CO2→NaHCO3

dd Y: NaHCO3

P2+Y

NaOH+NaHCO3→Na2CO3+H2O

dd Z: NaHCO3, Na2CO3

Z+Ca(NO3)2

Na2CO3+Ca(NO3)2→CaCO3+2NaNO3

2NaHCO3+Ca(NO3)2→CaCO3+2NaNO3+H2O+CO2