Question

cảm ơn ạaa

  

 

Answer 1

Câu 22:

\(n_{CH_3COOH}=\dfrac{45}{60}=0,75\left(mol\right)\)

PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄, t^o)

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,75\left(mol\right)\)

Mà: H = 80%
\(\Rightarrow n_{CH_3COOC_2H_5\left(TT\right)}=0,75.80\%=0,6\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(TT\right)}=0,6.88=52,8\left(g\right)\)

Answer 2

Câu 23:

m_CaCO3 = 20.80% = 16 (g) \(\Rightarrow n_{CaCO_3}=\dfrac{16}{100}=0,16\left(mol\right)\)

PT: \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

a, Theo PT: \(n_{CH_3COOH}=2n_{CaCO_3}=0,32\left(mol\right)\Rightarrow m_{CH_3COOH}=0,32.60=19,2\left(g\right)\)

b, \(n_{CO_2}=n_{CaCO_3}=0,16\left(mol\right)\Rightarrow V_{CO_2}=0,16.22,4=3,584\left(l\right)\)