Đáp án:
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Giải thích các bước giải:
Bài 1:
\(\begin{array}{l}
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
CaO + {H_2}O \to Ca{(OH)_2}\\
{K_2}O + {H_2}O \to 2KOH\\
BaO + {H_2}O \to Ba{(OH)_2}
\end{array}\)
\(\begin{array}{l}
b)\\
N{a_2}O + S{O_2} \to N{a_2}S{O_3}\\
CaO + S{O_2} \to CaS{O_3}\\
{K_2}O + S{O_2} \to {K_2}S{O_3}\\
BaO + S{O_2} \to BaS{O_3}
\end{array}\)
\(\begin{array}{l}
d)\\
N{a_2}O + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O\\
CaO + {H_2}S{O_4} \to CaS{O_4} + {H_2}O\\
CuO + {H_2}S{O_4} \to C{\rm{uS}}{O_4} + {H_2}O\\
MgO + {H_2}S{O_4} \to Mg{\rm{S}}{O_4} + {H_2}O\\
{K_2}O + {H_2}S{O_4} \to {K_2}S{O_4} + {H_2}O\\
BaO + {H_2}S{O_4} \to BaS{O_4} + {H_2}O\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
2F{e_3}{O_4} + 10{H_2}S{O_4} \to 3F{e_2}{(S{O_4})_3} + S{O_2} + 10{H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
ZnO + {H_2}S{O_4} \to ZnS{O_4} + {H_2}O
\end{array}\)
\(\begin{array}{l}
h)\\
N{a_2}O + 2HCl \to 2NaCl + {H_2}O\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{K_2}O + 2HCl \to 2KCl + {H_2}O\\
BaO + 2HCl \to BaC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
F{e_3}{O_4} + 8HCl \to 2FeC{l_3} + FeC{l_2} + 4{H_2}O\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O
\end{array}\)
\(\begin{array}{l}
i)\\
N{a_2}O + C{O_2} \to N{a_2}C{O_3}\\
CaO + C{O_2} \to CaC{O_3}\\
{K_2}O + C{O_2} \to {K_2}C{O_3}\\
BaO + C{O_2} \to BaC{O_3}
\end{array}\)
\(\begin{array}{l}
m)\\
CuO + {H_2} \to Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O
\end{array}\)
\(\begin{array}{l}
n)\\
CuO + C \to Cu + C{O_2}\\
ZnO + C \to Zn + C{O_2}\\
2F{e_2}{O_3} + 3C \to 4Fe + 3C{O_2}\\
F{e_3}{O_4} + 2C \to 3Fe + 2C{O_2}\\
2A{l_2}{O_3} + 9C \to 6CO + A{l_4}{C_3}
\end{array}\)
Bài 2:
\(\begin{array}{l}
1)\\
CaC{O_3} \to C{O_2} + {H_2}O\\
{n_{CaC{O_3}}} = 0,2mol\\
\to {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,2mol\\
{n_{NaOH}} = 0,28mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,28}}{{0,2}} = 1,4
\end{array}\)
=> Tạo 2 muối: \(N{a_2}C{O_3}\) và \(NaHC{O_3}\)
Gọi a và b lần lượt là số mol của \(C{O_2}\) (1) và \(C{O_2}\) (2)
\(\begin{array}{l}
C{O_2} + NaOH \to N{a_2}C{O_3} + {H_2}O(1)\\
C{O_2} + NaOH \to NaHC{O_3}(2)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
2a + b = 0,28
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,08\\
b = 0,12
\end{array} \right.\\
\to {n_{C{O_2}(1)}} = {n_{N{a_2}C{O_3}}} = 0,08mol\\
\to {n_{C{O_2}(2)}} = {n_{NaHC{O_3}}} = 0,12mol\\
\to {m_{N{a_2}C{O_3}}} = 8,48g\\
\to {m_{NaHC{O_3}}} = 10,08g
\end{array}\)
\(\begin{array}{l}
2)\\
{n_{C{O_2}}} = 0,2mol\\
{n_{Ca{{(OH)}_2}}} = 0,2mol\\
\to \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = 1
\end{array}\)
=> Tạo 1 muối: \(CaC{O_3}\)
\(\begin{array}{l}
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,2mol\\
\to {m_{CaC{O_3}}} = 20g
\end{array}\)
4,
\(\begin{array}{l}
a,\\
{n_{C{O_2}}} = 0,08mol\\
{n_{NaOH}} = 0,1mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 1,25\\
NaOH + C{O_2} \to NaHC{O_3}\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
\left\{ \begin{array}{l}
a + 2b = {n_{NaOH}} = 0,1mol\\
a + b = {n_{C{O_2}}} = 0,08mol
\end{array} \right.\\
\to a = 0,06mol \to b = 0,02mol\\
\to {n_{NaHC{O_3}}} = a = 0,06mol\\
\to {n_{N{a_2}C{O_3}}} = b = 0,02mol\\
\to {m_{NaHC{O_3}}} + {m_{N{a_2}C{O_3}}} = 7,16g
\end{array}\)
\(\begin{array}{l}
b,\\
{n_{C{O_2}}} = 0,12mol\\
{n_{NaOH}} = 0,1mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 0,83
\end{array}\)
Suy ra chỉ tạo ra 1 muối axit
\(\begin{array}{l}
NaOH + C{O_2} \to NaHC{O_3}\\
\to {n_{NaHC{O_3}}} = {n_{NaOH}} = 0,1mol\\
\to {m_{NaHC{O_3}}} = 8,4g
\end{array}\)
\(\begin{array}{l}
c,\\
{n_{C{O_2}}} = 0,04mol\\
{n_{NaOH}} = 0,1mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 2,5
\end{array}\)
Suy ra chỉ tạo 1 muối trung hòa
\(\begin{array}{l}
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
\to {n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,04mol\\
\to {m_{N{a_2}C{O_3}}} = 4,24g
\end{array}\)
