Lời giải:\(\begin{array}{l} \left. \begin{array}{l} {C_4}{H_8}{O_2}\\ {C_4}{H_6}{O_2}\;\;\;\\ {C_4}{H_4}{O_2} \end{array} \right\} \Rightarrow \;{C_4}{H_n}{O_2}\\ \begin{array}{*{20}{l}} \begin{array}{l} {d_{X/{O_2}}} = {\rm{ }}2,7\;{\rm{ }}\\ \Rightarrow {M_X}\; = {\rm{ }}86,4g \end{array}\\ { \Rightarrow {\rm{ }}n{\rm{ }} = {\rm{ }}6,4{\rm{ }}\left( {{C_4}{H_{6,4}}{O_2}} \right)} \end{array}\\ {n_X} = 0,015 \Rightarrow {n_{C{O_2}}} = 0,06mol,{n_{{H_2}O}} = 0,048mol\\ {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,06mol\\ \Rightarrow {m_{CaC{O_3}}} = 6g\\ \Rightarrow {m_{binhtang}} = {m_{C{O_2}}} + {m_{{H_2}O}} = 0,06.44 + 0,048.18 = 3,504(g) < {m_{CaC{O_3}}}\\ \Rightarrow {m_{dungdichgiam}} = 6 - 3,504 = 2,496g \end{array}\)
