Đáp án:

$\rm a)\\n_{(CH_3COO)_2Mg}=\dfrac{0,71}{142}=0,005\ (mol)$

Phản ứng xảy ra:

$\rm 2CH_3COOH+Zn\to (CH_3COO)_2Zn+H_2$

Theo phản ứng:

$\rm n_{CH_3COOH}=2n_{(CH_3COO)_2Zn}=0,01\ (mol)$

$\rm C_{M\ CH_3COOH}=\dfrac{0,01}{25×10^{-3}}=0,4\ (M)\\b)$

Phản ứng xảy ra:

$\rm CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo phản ứng:

$\rm n_{NaOH}=n_{CH_3COOH}=0,01\ (mol)$

$\rm m_{dd\ NaOH}=\dfrac{0,01×40}{10\%}=4\ (g)$