Đáp án:
\({m_{rắn}} = 8,8{\text{ gam}}\)
Giải thích các bước giải:
Gọi số mol \(CH_3COOC_2H_5;CH_3COOC_6H_5\) đều là \(x\)
\( \to 88x + 136y = 5,6 \to x = 0,025\)
Phản ứng xảy ra:
\(C{H_3}COO{C_2}{H_5} + NaOH\xrightarrow{{}}C{H_3}COONa + {C_2}{H_5}OH\)
\(C{H_3}COO{C_6}{H_5} + 2NaOH\xrightarrow{{}}C{H_3}COONa + {C_6}{H_5}ONa + {H_2}O\)
\({n_{NaOH}} = 0,1.1,2 = 0,12{\text{ mol > }}{{\text{n}}_{C{H_3}COO{C_2}{H_5}}} + 2{n_{C{H_3}COO{C_6}{H_5}}}\)
\( \to {n_{{C_2}{H_5}OH}} = {n_{C{H_3}COO{C_2}{H_5}}} = 0,025{\text{ }}mol;{n_{{H_2}O}} = {n_{C{H_3}COO{C_6}{H_5}}} = 0,025{\text{ mol}}\)
BTKL:
\({m_X} + {m_{NaOH}} = {m_{rắn}} + {m_{{C_2}{H_5}OH}} + {m_{{H_2}O}}\)
\( \to 5,6 + 0,12.40 = {m_{rắn}} + 0,025.46 + 0,025.18\)
\( \to {m_{rắn}} = 8,8{\text{ gam}}\)
