Câu 2:

$\rm a)\ C_2H_5OH+O_2\xrightarrow{t^°,\ men}CH_3COOH+H_2O\\b)\ C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2\\c)\ 2CH_3COOH+Fe\to (CH_3COO)_2Fe+H_2O\\d)\ (RCOO)_3C_3H_5+3NaOH\xrightarrow{t^°}3RCOONa+C_3H_5(OH)_3$

Câu 3:

$\rm a)$

Phản ứng xảy ra:

$\rm Mg+2CH_3COOH\to (CH_3COO)_2Mg+H_2\\ZnO+2CH_3COOH\to (CH_3COO)_2Zn+H_2O$

$\rm b)\\n_{H_2}=\dfrac{3,36}{22,4}=0,15\ (mol)$

Theo phương trình:

$\rm n_{Mg}=n_{H_2}=0,15\ (mol)$

Khối lượng mỗi chất:

$\rm m_{Mg}=0,15×24=3,6\ (g)\\m_{ZnO}=11,7-3,6=8,1\ (g)$

$\rm c)\\n_{ZnO}=\dfrac{8,1}{81}=0,1\ (mol)$

Theo phương trình:

$\rm n_{CH_3COOH}=2×(n_{Mg}+n_{ZnO})=0,5\ (mol)$

$\rm C\%_{CH_3COOH}=\dfrac{0,5×60}{200}×100\%=15\%$

$\rm d)$

Theo phương trình:

$\rm n_{(CH_3COO)_2Mg}=n_{Mg}=0,15\ (mol)\\n_{(CH_3COO)_2Zn}=n_{ZnO}=0,1\ (mol)$

Bảo toàn khối lượng:

$\rm m_{dd\ sau\ pu}=11,7+200-0,15×2=211,4\ (g)$

$\rm C\%_{(CH_3COO)_2Mg}=\dfrac{0,15×142}{211,4}×100\%=10,08\%\\C\%_{(CH_3COO)_2Zn}=\dfrac{0,1×183}{211,4}×100\%=8,66\%$