Đáp án:
\({V_{{C_2}{H_2}}}{\text{ = }}{{\text{V}}_{C{H_4}}} = 0,168{\text{ lít}}\)
\({m_{CaC{O_3}}} = 2,25{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)
Ta có:
\({n_{B{r_2}}} = \frac{{2,4}}{{80.2}} = 0,015{\text{ mol}} \to {{\text{n}}_{{C_2}{H_2}}} = \frac{1}{2}{n_{B{r_2}}} = 0,0075{\text{ mol}}\)
\( \to {V_{{C_2}{H_2}}} = 0,0075.22,4 = 0,168\;lí{\text{t = }}{{\text{V}}_{C{H_4}}}{\text{ }}\)
Đốt cháy hỗn hợp \(X\)
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
\(2{C_2}{H_2} + 5{O_2}\xrightarrow{{{t^o}}}4C{O_2} + 2{H_2}O\)
\( \to {n_{C{O_2}}} = {n_{C{H_4}}} + 2{n_{{C_2}{H_2}}} = 0,0075 + 0,0075.2 = 0,0225{\text{ mol}}\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
\( \to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,0225{\text{ mol}}\)
\( \to {m_{CaC{O_3}}} = 0,0225.100 = 2,25{\text{ gam}}\)
