Đáp án:
\( {V_{{O_2}}} = 3,78{\text{ lít}}\)
\({m_{{H_2}O}} = 2,7{\text{gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_n}{H_{2n + 1}}OH + 1,5n{O_2}\xrightarrow{{{t^o}}}nC{O_2} + (n + 1){H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{2,52}}{{22,4}} = 0,1125{\text{ mol}}\)
\( \to {n_{ancol}} = \frac{{{n_{C{O_2}}}}}{n} = \frac{{0,1125}}{n}\)
\( \to {M_{ancol}} = 14n + 18 = \frac{{2,25}}{{\frac{{0,1125}}{n}}} = 20n \to n = 3\)
Vậy \(A\) là \(C_3H_7OH\)
Ta có:
\({n_A} = \frac{{2,25}}{{60}} = 0,0375{\text{ mol}}\)
\( \to {n_{{O_2}}} = 1,5n.{n_A} = 0,16875{\text{ mol}}\)
\({n_{{H_2}O}} = (n + 1).{n_A} = 0,15{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,16875.22,4 = 3,78{\text{ lít}}\)
\({m_{{H_2}O}} = 0,15.18 = 2,7{\text{gam}}\)
