Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
Bài 1:
a,
Oxit bazo:
\({K_2}O\): Kali oxit
\(F{e_2}{O_3}\): Sắt (III) oxit
Oxit axit:
\(S{O_3}\): lưu huỳnh trioxit
\({P_2}{O_5}\): Điphotpho pentaoxit
b,
\(\begin{array}{l}
{K_2}O + {H_2}O \to 2K{\rm{OH}}\\
{\rm{S}}{{\rm{O}}_3} + {H_2}O \to {H_2}S{O_4}\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
{K_2}O + 2HCl \to 2KCl + {H_2}O\\
F{{\rm{e}}_2}{O_3} + 6HCl \to 2F{\rm{e}}C{l_3} + 3{H_2}O\\
2K{\rm{O}}H + S{O_3} \to {K_2}S{O_4} + {H_2}O\\
6K{\rm{O}}H + {P_2}{O_5} \to 2{K_3}P{O_4} + 3{H_2}O
\end{array}\)
Bài 2:
CaO:
\(\begin{array}{l}
CaO + {H_2}O \to Ca{(OH)_2}\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CaO + S{O_3} \to CaS{O_4}
\end{array}\)
\(S{O_2}\):
\(\begin{array}{l}
S{O_2} + {H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H_2}S{O_3}\\
S{O_2} + BaO \to BaS{O_3}\\
S{O_2} + Ba{(OH)_2} \to BaS{O_3} + {H_2}O
\end{array}\)
Bài 3:
\(\begin{array}{l}
F{\rm{e}} + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
\to {m_{Fe}} = 8,4g\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,15mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,15}}{{0,05}} = 3M
\end{array}\)