1)
Hợp chất hữu cơ \(A\) có dạng \(C_xH_y\)
Đốt cháy \(A\)
\({C_x}{H_y} + (x + \frac{y}{4}){O_2}\xrightarrow{{{t^o}}}xC{O_2} + \frac{y}{2}{H_2}O\)
Ta có:
\({n_{{H_2}O}} = \frac{{1,44}}{{18}} = 0,08{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,08.2 = 0,16{\text{ mol}}\)
\( \to y = \frac{{{n_{{H_2}O}}}}{{{n_A}}} = \frac{{0,16}}{{0,02}} = 8\)
\(x + \frac{y}{4} = \frac{{{n_{{O_2}}}}}{{{n_A}}} = \frac{{0,1}}{{0,02}} = 5 \to x = 3\)
Vậy \(A\) là \(C_3H_8\)
2)
Phản ứng xảy ra:
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{{t^o}}}CaC{O_3} + {H_2}O\)
\({n_{C{H_4}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol}}\)
\( \to {n_{{O_2}}} = 2{n_{C{H_4}}} = 0,05.2 = 0,1{\text{ mol}}\)
\( \to {m_{{O_2}}} = 0,1.32 = 3,2{\text{ gam}}\)
\({n_{C{O_2}}} = {n_{C{H_4}}} = 0,05{\text{ mol}}\)
\( \to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,05{\text{ mol}}\)
\( \to {m_{CaC{O_3}}} = 0,05.100 = 5{\text{ gam}}\)
