Đáp án:

 A

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_3}{H_6}{O_2} + \frac{7}{2}{O_2}\xrightarrow{{{t^o}}}3C{O_2} + 3{H_2}O\)

\({C_3}{H_4}{O_2} + 3{O_2}\xrightarrow{{{t^o}}}3C{O_2} + 2{H_2}O\)

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

Ta có: 

\({n_{C{O_2}}} = 3{n_{{C_3}{H_6}{O_2}}} + 3{n_{{C_3}{H_4}{O_2}}} = 3.0,3 = 0,9{\text{ mol}}\)

\( \to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,9{\text{ mol}}\)

Gọi: \({n_{{H_2}O}} = x\)

\( \to {m_{dd{\text{ giảm}}}} = {m_{CaC{O_3}}} - {m_{C{O_2}}} - {m_{{H_2}O}} = 37,8{\text{ gam}} \to {\text{0}}{\text{,9}}{\text{.100 - 0}}{\text{,9}}{\text{.44 - 18x = 37}}{\text{,8}} \to {\text{x = 0}}{\text{,7}} \to {{\text{m}}_{{H_2}O}} = 0,7.18 = 12,6{\text{ gam}}\)