1)
Hidrocacbon có dạng \(C_xH_y\)
\( \to {M_A} = 12x + y = 0,5{M_{{O_2}}} = 0,5.32 = 16\)
\( \to \% {m_C} = \frac{{12x}}{{{M_A}}} = \frac{{12x}}{{16}} = 75\% \to x = 1 \to y = 4\)
Vậy \(A\) là \(CH_4\)
2)
Đốt cháy \(C_2H_2\)
\(2{C_2}{H_2} + 5{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 2{H_2}O\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
Ta có:
\({V_{{O_2}}} = \frac{5}{2}{V_{{C_2}{H_2}}} = 8,4{\text{ lít}}\)
\( \to {V_{kk}} = \frac{{8,4}}{{21\% }} = 40{\text{ lít}}\)
\({n_{{C_2}{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{n}}_{C{O_2}}} = 2{n_{{C_2}{H_2}}} = 0,3{\text{ mol = }}{{\text{n}}_{CaC{O_3}}}\)
\( \to m = {m_{CaC{O_3}}} = 0,3.100 = 30{\text{ gam}}\)
