Đáp án đúng: A
8,00g.
${C}{{{O}}_{{2}}}$ phản ứng với$\displaystyle {Ca}{{\left( {OH} \right)}_{{2}}}$ tạo 5 gam${CaC}{{{O}}_{{3}}}$và${Ca(HC}{{{O}}_{{3}}}{{{)}}_{{2}}}$
$\begin{array}{l}\xrightarrow{BTNT\,\,Ca}\,{{n}_{Ca{{(HC{{O}_{3}})}_{2}}}}={{n}_{Ca{{(OH)}_{2}}}}-{{n}_{CaC{{O}_{3}}}}=0,1-0,05=0,05\\\xrightarrow{BTNT\,C}\,{{n}_{C{{O}_{2}}}}={{n}_{CaC{{O}_{3}}}}+2{{n}_{Ca{{(HC{{O}_{3}})}_{2}}}}=0,15\,mol\\\Rightarrow {{{n}}_{{O}\,\,{(oxit)}}}\,{=}\,{{{n}}_{{C}{{{O}}_{{2}}}}}\,{=}\,{0,15}\,{mol}\,\Rightarrow \,{{{n}}_{{C}{{{l}}^{{-}}}}}\,{=}\,{2}{{{n}}_{{O}\,}}{=}\,{0,3}\,{mol}\\\Rightarrow \,{{{m}}_{{Fe}}}{=}\,{16,25}\,{-}\,{0,3}{.35,5}\,{=}\,\,5,6\,{gam}\Rightarrow \frac{{y}}{{x}}{=}\frac{{{{n}}_{{O}}}}{{{{n}}_{{Fe}}}}{=1,5}\in \,{ }\!\![\!\!{ 1;}\frac{{3}}{{2}}{ }\!\!]\!\!{ }\\\Rightarrow {m}\,{=}\,{{{m}}_{{Fe}}}\,{+}\,{{{m}}_{{O}}}=5,6\,+\,0,15.16\,=\,8\end{array}$