Đáp án:
D
Giải thích các bước giải:
\(\begin{array}{l}
{M_x}{O_y} + 2yHCl \to xMC{l_{\frac{{2y}}{x}}} + y{H_2}O\\
nHCl = 0,2 \times 1,5 = 0,3\,mol\\
nMC{l_{\dfrac{{2y}}{x}}} = \dfrac{{0,3 \times x}}{{2y}} = \dfrac{{0,15x}}{y}\,mol\\
MMC{l_{\dfrac{{2y}}{x}}} = \dfrac{{19,05}}{{\dfrac{{0,15x}}{y}}} = \dfrac{{127y}}{x}\,g/mol\\
= > MM + \dfrac{{35,5 \times 2y}}{x} = \dfrac{{127y}}{x}\\
= > MM = 28 \times \dfrac{{2y}}{x}\,g/mol\\
\dfrac{{2y}}{x} = 2 = > MM = 56g/mol = > M:Fe\\
\dfrac{{2y}}{x} = 2 = > x = y\\
= > CTHH:FeO\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
nFeO = \dfrac{{0,3}}{2} = 0,15\,mol\\
mFeO = 0,15 \times 72 = 10,8g
\end{array}\)