Đáp án:
$\%m_C=19,67\%$
$\%m_H=4,92\%$
$\%m_N=22,95\%$
$\%m_O=52,46\%$
Giải thích các bước giải:
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O + {N_2}\)
Ta có:
\({n_{C{O_2}}} = \dfrac{{0,44}}{{44}} = 0,01{\text{ mol = }}{{\text{n}}_C}\)
\({n_{{H_2}O}} = \dfrac{{0,27}}{{18}} = 0,015{\text{ mol}} \\\to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,03{\text{ mol}}\)
\({n_{{N_2}}} = \dfrac{{0,112}}{{22,4}} = 0,005 \\\to {n_N} = 2{n_{{N_2}}} = 0,01{\text{ mol}}\)
\( \to {m_C} = 0,01.12 = 0,12{\text{ gam;}}\\{{\text{m}}_H} = 0,03.1 = 0,03{\text{ gam;}}\\{{\text{m}}_N} = 0,01.14 = 0,14{\text{ gam}}\)
\( \to {m_O} = 0,61 - 0,12 - 0,03 - 0,14 = 0,32{\text{ gam}}\)
\( \to \% {m_C} = \dfrac{{0,12}}{{0,61}} = 19,67\% ;\\\% {m_H} = \dfrac{{0,03}}{{0,61}} = 4,92\% ;\\\% {m_N} = \dfrac{{0,14}}{{0,61}} = 22,95\% \)
\( \to \% {m_O} = 52,46\% \)
