$22)$

$m_{(C_6H_{10}O_5)_n}=1.6 \times 95\% \times 10^3(gam)$

`=>`$n_{(C_6H_{10}O_5)_n}=\dfrac{1.6 \times 95\% \times 10^3}{162n}(mol)$

$(C_6H_{10}O_5)_n \to nC_6H_{12}O_6 \to 2nCO_2+2nC_2H_5OH$

`->`$n_{C_2H_5OH}=2n \times n_{(C_6H_{10}O_5)_n} \times \dfrac{H}{100}$

$=2n \times \dfrac{1.6 \times 95\% \times 10^3}{162n} \times  \dfrac{H}{100}(mol)$

`=>`$m_{C_2H_5OH}=46 \times 2 \times \dfrac{1.6 \times 95\% \times 10^3}{162} \times  \dfrac{H}{100}(gam)$ $(1)$

$\dfrac{V_{C_2H_5OH}}{V_r} \times 100\%=35$

`=>`$V_{C_2H_5OH}=0.35 \times 2(l) =0.35 \times 2 \times 10^3(mL)$

`=>`$m_{C_2H_5OH}=V_{C_2H_5OH} \times d=0.35 \times 2 \times 10^3 \times 0.8(gam)$ $(2)$

Từ $(1)$ và $(2)$ suy ra : $H\%≈64.87\%$

Đáp án:

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