Đáp án:

Câu 11:

$\rm a)$

Phương trình phản ứng:

$\rm C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2\\H_2O+Na\to NaOH+\dfrac{1}{2}H_2$

$\rm b)\\V_{ruou\ nguyen\ chat}=20×96\%=19,2\ (mL)\\m_{ruou\ nguyen\ chat}=19, 2×0,8=15,36\ (g)$

$\rm c)\\V_{H_2O}=20-19,2=0,8\ (mL)\\n_{H_2O}=\dfrac{0,8}{18}=\dfrac{2}{45}\ (mol)\\n_{C_2H_5OH}=\dfrac{15,36}{46}=\dfrac{192}{575}\ (mol)$

Theo phương trình:

$\rm n_{H_2}=\dfrac{1}{2}×(n_{C_2H_5OH}+n_{H_2O})=\dfrac{979}{5175}\ (mol)$

$\rm V_{H_2}=\dfrac{979}{5175}×22,4=4,24\ (L)$