Đáp án:
\({C_3}{H_8}\)
Giải thích các bước giải:
X có dạng \({C_x}{H_y}\)
\({C_x}{H_y} + (x + \frac{y}{4}){O_2}\xrightarrow{{}}xC{O_2} + \frac{y}{2}{H_2}O\)
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
\({m_{BaC{O_3}}} = 29,55{\text{ gam}} \to {{\text{n}}_{BaC{O_3}}} = \frac{{29,55}}{{197}} = 0,15{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)
\(\to {m_{dd{\text{ giảm}}}} = {m_{BaC{O_3}}} - ({m_{C{O_2}}} + {m_{{H_2}O}}) \to {m_{C{O_2}}} + {m_{{H_2}O}} = 29,55 - 19,35 = 10,2{\text{ gam}}\)
\(\to {m_{{H_2}O}} = 10,2 - 0,15.44 = 3,6{\text{ gam}} \to {{\text{n}}_{{H_2}O}} = \frac{{3,6}}{{18}} = 0,2{\text{ mol > }}{{\text{n}}_{C{O_2}}}\)
Vậy X là ankan
\(\to {n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,2 - 0,15 = 0,05{\text{ mol}} \to {{\text{C}}_X} = \frac{{0,15}}{{0,05}} = 3 \to X:{C_3}{H_8}\)
