Đáp án:
a) 30g
b) 15g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
n{C_2}{H_5}OH = \dfrac{m}{M} = \dfrac{{6,9}}{{46}} = 0,15\,mol\\
nC{O_2} = 2n{C_2}{H_5}OH = 0,3\,mol\\
nCaC{O_3} = nC{O_2} = 0,3\,mol\\
mCaC{O_3} = n \times M = 0,3 \times 100 = 30g\\
b)\\
{C_6}{H_{12}}{O_6} \to 2{C_2}{H_5}OH + 2C{O_2}\\
n{C_6}{H_{12}}{O_6} = \dfrac{{0,15}}{2} = 0,075\,mol\\
H = 90\% \Rightarrow m{C_6}{H_{12}}{O_6} = \dfrac{{0,075 \times 180}}{{90\% }} = 15g
\end{array}\)