a, Ta có: \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
PT: \(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
________0,2______0,6 (mol)
\(\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
b, Ta có: VC2H5OH = 9,2:0,8 = 11,5 (ml)
⇒ \(8=\dfrac{11,5}{V_{rượu8^o}}.100\)
⇒ V rượu 8 độ = 143,75 (ml)
c, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=0,2\left(mol\right)\)
Mà: H = 80%
\(\Rightarrow n_{CH_3COOH\left(TT\right)}=0,2.80\%=0,16\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=0,16.60=9,6\left(g\right)\)