Đáp án:
a) $V_{H_2}=4,48\ (l)$
b) $m_{dd\ CH_3COOH}=120\ (g)$
Giải thích các bước giải:
PTHH: $Zn + 2CH_3COOH \to Zn(CH_3COO)_2 + H_2↑$
a) Ta có: $n_{Zn}=\dfrac{13}{65}=0,2\ (mol)$
Theo PTHH: $n_{H_2↑} = n_{Zn} = 0,2\ (mol)$
$⇒ V_{H_2}=0,2.22,4=4,48\ (l)$
b)
Theo PTHH: $n_{CH_3COOH}=2n_{Zn}=2.0,2=0,4\ (mol)$
$ m_{CH_3COOH}=n.M=0,4.60=24\ (g)$
Mặt khác: $C\% = \dfrac{m_{ct}}{m_{dd}}.100\%$
$⇒m_{dd}=\dfrac{m_{ct}.100\%}{C\%}$
$⇒m_{dd\ CH_3COOH}= \dfrac{24.100}{20}=120\ (g)$