Đáp án:
b) 3,08l
c) 20,075 g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Cu + {O_2} \to 2CuO\\
2Fe + {O_2} \to 2FeO\\
4Fe + 3{O_2} \to 2F{e_2}{O_3}\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}\\
F{e_3}{O_4} + 8HCl \to 2FeC{l_3} + FeC{l_2} + 4{H_2}O\\
b)\\
m{O_2} = mhhX - mhh = 19,2 - 14,8 = 4,4\,g\\
n{O_2} = \frac{{4,4}}{{32}} = 0,1375\,mol\\
V{O_2} = 0,1375 \times 22,4 = 3,08l\\
c)\\
nO = 0,1375 \times 2 = 0,275\,mol\\
nHCl = 2nO = 0,55\,mol\\
mHCl = 0,55 \times 36,5 = 20,075g
\end{array}\)