PTHH:`2CH_3COOH+Zn→(CH_3COO)_2Zn+H_2↑` (1)
`a)`
Ta có: `m_(Zn)=9,75/65=0,15`(mol)
Theo `PTHH`,ta thấy: `n_(H_2)=n_Zn=0,15(mol)`
`⇒V_(H_2)=0,15.22,4=3,36`(l)
`b)`
Theo `PTHH`,ta thấy: `n_(H_2)=n_((CH_3COO)_2Zn)=0,15`(mol)
`⇒m_((CH_3COO)_2Zn)=0,15.(12+3+12+16.2)=8,85`(g)
`c)`
`PTHH: C_2H_5OH \overset{Men -giấm}\to CH_3COOH+H_2O`(2)
Theo `PTHH`(1),ta thấy: `n_(CH_3COOH)=2n_(H_2)=2.0,15=0,3`(mol)
Theo `PTHH(2)`,ta thấy: `n_(CH_3COOH)=n_(C_2H_5OH)=0,3`(mol)
`⇒m_(ctC_2H_5OH)=n.M=0,3.46=13,8 (g)`
Mà: `%C_2H_5OH=(m_(ctC_2H_5OH))/(m_(ddC_2H_5OH)).100`
`⇒m_(ddC_2H_5OH)=100.(13,8)/20=69`(g)
Lại có hiệu suất đạt `80%` nên,ta có:
$⇒ m_{ttC_2H_5OH}=\dfrac{ H × m_{ltC_2H_5OH}}{100\%}= \dfrac{ 80 × 69 }{100}=55,2$ (g)
Xin hay nhất =_=