Đáp án:
\({C_3}{H_8}{O_3}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(X + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Hấp thụ sản phẩm cháy vào 0,1 mol \(Ba{(OH)_2}\)
TH1: \(Ba{(OH)_2}\) dư
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
\({n_{BaC{O_3}}} = \frac{{9,85}}{{197}} = 0,05{\text{ mol = }}{{\text{n}}_{C{O_2}}} \to {m_{C{O_2}}} = 0,05.44 = 2,2{\text{ gam}} \to {{\text{m}}_{{H_2}O}} = \frac{{2,2}}{{11}}.6 = 1,2{\text{ gam}} \to {{\text{n}}_{{H_2}O}} = \frac{1}{2}18 = \frac{1}{{15}}{\text{ mol}}\)
Bảo toàn nguyên tố:
\({n_C} = {n_{C{O_2}}} = 0,05{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = \frac{2}{{15}}{\text{ mol}} \to {{\text{n}}_O} = \frac{{4,6 - 0,05.12 - \frac{2}{{15}}.1}}{{16}} = \frac{{29}}{{120}}\)
\( \to {n_C}:{n_H}:{n_O} = 0,05:\frac{2}{{15}}:\frac{{29}}{{120}} = 6:16:29\)
\( \to {C_6}{H_{16}}{O_{29}}\) (không tồn tại)
TH2: \(Ba{(OH)_2}\) hết
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
\(Ba{(OH)_2} + 2C{O_2}\xrightarrow{{}}Ba{(HC{O_3})_2}\)
\({n_{BaC{O_3}}} = 0,05{\text{ mol}} \to {{\text{n}}_{Ba{{(HC{O_3})}_2}}} = 0,1 - 0,05 = 0,05{\text{ mol}} \to {{\text{n}}_{C{O_2}}} = {n_{BaC{O_3}}} + 2{n_{Ba{{(HC{O_3})}_2}}} = 0,15{\text{ mol}}\)
\(\to {m_{C{O_2}}} = 0,15.44 = 6,6{\text{ gam}} \to {{\text{m}}_{{H_2}O}} = \frac{{6,6}}{{11}}.6 = 3,6{\text{ gam}} \to {{\text{n}}_{{H_2}O}} = \frac{{3,6}}{{18}} = 0,2{\text{ mol}}\)
Bảo toàn nguyên tố:
\({n_C} = {n_{C{O_2}}} = 0,15{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,4{\text{ mol}} \to {{\text{n}}_O} = \frac{{4,6 - 0,15.12 - 0,4}}{{16}} = 0,15{\text{ mol}}\)
\( \to {n_C}:{n_H}:{n_O} = 0,15:0,4:0,15 = 3:8:3\)
CTĐGN của Y là \({C_3}{H_8}{O_3}\)
