Đáp án:

${V_{{O_2}}} = 2,912(l)$

$\% {m_{C{H_3}OH}} = 16,2\% $

Giải thích các bước giải:

${m_{O(X)}} = {m_X}.\dfrac{{54,6835}}{{100}} = 2,16g \Rightarrow {n_{O(X)}} = 0,135mol$ 

${n_C} = {n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{12,5}}{{100}} = 0,125mol$

${m_X} = {m_O} + {m_C} + {m_H} \Rightarrow {m_H} = 3,95 - 2,16 - 0,125.12 = 0,29g$

$ \Rightarrow {n_H} = 0,29mol \Rightarrow {n_{{H_2}O}} = \dfrac{1}{2}{n_H} = 0,145mol$

Bảo toàn nguyên tố $O$: ${n_{O(X)}} + 2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}$

$ \Rightarrow {n_{{O_2}}} = \dfrac{{0,125.2 + 0,145 - 0,135}}{2} = 0,13mol$

$ \Rightarrow {V_{{O_2}}} = 0,13.22,4 = 2,912(l)$

${n_{C{H_3}OH}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,145 - 0,125 = 0,02mol$

$ \Rightarrow \% {m_{C{H_3}OH}} = \dfrac{{0,02.32}}{{3,95}}.100\%  = 16,2\% $