Đáp án:
a) 2,24l
b) 5,0625 g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
n{C_2}{H_5}OH = \dfrac{m}{M} = \dfrac{{2,3}}{{46}} = 0,05\,mol\\
nC{O_2} = 2n{C_2}{H_5}OH = 0,1\,mol\\
VC{O_2} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
b)\\
{({C_6}{H_{10}}{O_5})_n} + n{H_2}O \to n{C_6}{H_{12}}{O_6}\\
{C_6}{H_{12}}{O_6} \to 2{C_2}{H_5}OH + 2C{O_2}\\
n{C_6}{H_{12}}{O_6} = \dfrac{{0,05}}{2} = 0,025\,mol\\
n{({C_6}{H_{10}}{O_5})_n} = \dfrac{{0,025}}{n}\,mol\\
m{({C_6}{H_{10}}{O_5})_n} = \dfrac{{0,025}}{n} \times 162n = 4,05g
\end{array}\)
\(H = 80\% \Rightarrow m = \dfrac{{4,05}}{{80\% }} = 5,0625g\)
