Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

4,

a,

\(\begin{array}{l}
S + {O_2} \to S{O_2}\\
S{O_2} + \dfrac{1}{2}{O_2} \to S{O_3}\\
2S{O_3} + Ca{(OH)_2} \to Ca{(HS{O_4})_2}\\
Ca{(HS{O_4})_2} + 2NaHS{O_3} \to CaS{O_4} + N{a_2}S{O_4} + 2S{O_2} + 2{H_2}O
\end{array}\)

b,

\(\begin{array}{l}
CaO + {H_2}O \to Ca{(OH)_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
CaC{O_3} \to CaO + C{O_2}\\
NaOH + C{O_2} \to NaHC{O_3}
\end{array}\)

5,

\(\begin{array}{l}
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{CuO}} = 0,04mol\\
 \to {n_{CuS{O_4}}} = {n_{CuO}} = 0,04mol\\
 \to {m_{CuS{O_4}}} = 6,4g
\end{array}\)

6,

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,9mol\\
\dfrac{{{n_{Al}}}}{2} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3}
\end{array}\)

Suy ra dung dịch sau phản ứng có \(A{l_2}{(S{O_4})_3}\) và \({H_2}S{O_4}\) dư

\(\begin{array}{l}
 \to {n_{{H_2}S{O_4}}}dư= 0,9 - \dfrac{3}{2}{n_{Al}} = 0,6mol\\
 \to {n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,1mol\\
 \to {m_{{H_2}S{O_4}}}dư= 58,8g\\
 \to {m_{A{l_2}{{(S{O_4})}_3}}} = 34,2g
\end{array}\)

\(\begin{array}{l}
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
 \to {V_{{H_2}}} = 6,72l
\end{array}\)

7,

\(\begin{array}{l}
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = 0,15mol\\
 \to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15mol\\
 \to {m_{CaC{O_3}}} = 15g
\end{array}\)

8,

\(\begin{array}{l}
MgO + 2HN{O_3} \to Mg{(N{O_3})_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
S{O_3} + {K_2}O \to {K_2}S{O_4}\\
CaS{O_3} + 2HCl \to CaC{l_2} + S{O_2} + {H_2}O\\
K + {H_2}O \to KOH + \dfrac{1}{2}{H_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
Ca{(HC{O_3})_2} \to CaC{O_3} + C{O_2} + {H_2}O
\end{array}\)