Đáp án + Giải thích các bước giải:

$a\big)$

$C_6H_{12}O_6\xrightarrow[men\,rượu]{30-32^oC}2C_2H_5OH+2CO_2$

$CO_2+Ca(OH)_2\to CaCO_3↓+H_2O$

$b\big)$

Theo PT: $n_{C_2H_5OH}=n_{CaCO_3}=\frac{20}{100}=0,2(mol)$

$\to m_{C_2H_5OH}=0,2.46=9,2(g)$

$c\big)$

Theo PT: $n_{C_6H_{12}O_6(tt)}=\frac{1}{2}n_{C_2H_5OH}=0,1(mol)$

$\to a=m_{C_6H_{12}O_6(lt)}=\frac{0,1.180}{80\%}=22,5(g)$

Đáp án:

\({m_{{C_2}{H_5}OH}} = 9,2{\text{ gam}}\)

\(  {\text{a = }}{{\text{m}}_{{C_6}{H_{12}}{O_6}}}  = 22,5{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{{\text{O}}_3} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{20}}{{100}} = 0,2{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}OH}}\)

\( \to {m_{{C_2}{H_5}OH}} = 0,2.46 = 9,2{\text{ gam}}\)

\({n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}} = \frac{1}{2}{n_{C{O_2}}} = 0,1{\text{ mol}}\)

\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{1}{{80\% }} = 0,125{\text{ mol}} \to {\text{a = }}{{\text{m}}_{{C_6}{H_{12}}{O_6}}} = 0,125.180 = 22,5{\text{ gam}}\)