Question

Answer 1

a) \(V_{C_2H_5OH\left(bđ\right)}=\dfrac{5.200000}{100}=10^4\left(l\right)\)

=> \(m_{C_2H_5OH\left(bđ\right)}=10^4.0,8=8000\left(kg\right)=8.10^6\left(g\right)\)

=> \(n_{C_2H_5OH\left(bđ\right)}=\dfrac{8.10^6}{46}=\dfrac{4.10^6}{23}\left(mol\right)\)

=> \(n_{C_2H_5OH\left(pư\right)}=\dfrac{4.10^6}{23}.90\%=\dfrac{18}{115}.10^6\left(mol\right)\)

PTHH: C₂H₅OH + O₂ --men giấm--> CH₃COOH + H₂O

=> \(n_{CH_3COOH}=\dfrac{18}{115}.10^6\left(mol\right)\)

=> \(m_{CH_3COOH}=\dfrac{18}{115}.10^6.60=\dfrac{216}{23}.10^6\left(g\right)=\dfrac{216}{23}\left(tấn\right)\)

b) 

\(m_{dd.CH_3COOH}=\dfrac{\dfrac{216}{23}.100}{5}=\dfrac{4320}{23}\left(tấn\right)\)

Answer 2

cái này ở solvve thì phẻi=))