Đáp án:
CHÚC BẠN HỌC TỐT !!!!!!!!
Giải thích các bước giải:
$m CH_3COOH = 200.10$% $= 20 (g)$
$=> n CH_3COOH = \frac{20}{60} = \frac{1}{3} (mol)$
$PTHH:$
$CH_3COOH + NaOH → CH_3COOH + H_2O$
$a)$
$Theo$ $PTHH:$
$n NaOH = n CH_3COOH = \frac{1}{3} (mol)$
$=> m NaOH = \frac{1}{3}.40 = \frac{40}{3} (g)$
$b)$
$Theo$ $PTHH:$
$n CH_3COONa = \frac{1}{3} (mol)$
$=> m CH_3COONa = \frac{1}{3} . 82 = \frac{82}{3} (g)$
$Ta$ $có:$
$m dd = 200 + \frac{40}{3} : 8$% $= \frac{1100}{3} (g)$
$=> C$% $= \frac{CH_3COOH}{m dd}.100% = \frac{\frac{82}{3}}{\frac{1100}{3}}.100$% $≈ 7,45$%
