Đáp án:
b) 3,08l
c) 20,075g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
4Fe + 3{O_2} \xrightarrow{t^0} 2F{e_2}{O_3}\\
2Fe + {O_2} \xrightarrow{t^0} 2FeO\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
F{e_3}{O_4} + 8HCl \to 2FeC{l_3} + FeC{l_2} + 4{H_2}O\\
b)\\
BTKL\,:{m_{hh}} + {m_{{O_2}}} = {m_{hhX}} \Rightarrow {m_{{O_2}}} = {m_{hhX}} - {m_{hh}} = 19,2 - 14,8 = 4,4g\\
{n_{{O_2}}} = \dfrac{{4,4}}{{32}} = 0,1375\,mol\\
{V_{{O_2}}} = 0,1375 \times 22,4 = 3,08l\\
c)\\
{n_O} = 2{n_{{O_2}}} = 0,275\,mol\\
{n_{HCl}} = 2{n_O} = 0,55\,mol\\
{m_{HCl}} = 0,55 \times 36,5 = 20,075g
\end{array}\)
