Đáp án:
\(m = 1{\text{ gam}}\)
\(\% {m_C} = 90\% \to \% {m_H} = 10\% \)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
Ta có:
\({n_{CaC{O_3}}} = \frac{{7,5}}{{100}} = 0,075{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)
\({m_{bình{\text{ tăng}}}} = {m_{C{O_2}}} + {m_{{H_2}O}} = 0,075.44 + {m_{{H_2}O}} = 4,2\)
\( \to {m_{{H_2}O}} = 0,9{\text{ gam}}\)
\( \to {n_{{H_2}O}} = \frac{{0,9}}{{18}} = 0,05{\text{ mol}}\)
\({n_{{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
\({n_C} = {n_{C{O_2}}} = 0,075{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,1{\text{ mol}}\)
\({n_O} = 0,075.2 + 0,05 - 0,1.2 = 0\)
\(A\) chỉ chứa \(C;H\)
\( \to m = {m_C} + {m_H} = 0,075.12 + 0,1.1 = 1{\text{ gam}}\)
\(\% {m_C} = \frac{{0,075.12}}{1} = 90\% \to \% {m_H} = 10\% \)
