oxide: Na2O: sodium oxide

            ZnO: zinc oxide

               Fe2O3: iron(II) oxide

               N2O3: nitrogen oxide 

    acid: HNO3: hydrogen nitrate

              HCl :  hydro chloric acid

             H2S : hydro sulfua

                    H2SO4: hydrogen sulfate

base: Ba(OH)2 : barium hydroxide

          Al(OH)3: aluminium hydroxide

 muoi:   Al(NO3)3: aluminium nitrate

              CaCO3: calcium carbonate

\(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

_______________0,15______0,05 (mol)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

Giúp mình với

a, \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

b, Ta có: \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=1.60=60\left(g\right)\)

\(\Rightarrow m_{C_2H_5OH}=90-60=30\left(g\right)\)

c, \(n_{C_2H_5OH}=\dfrac{30}{46}=\dfrac{15}{23}\left(mol\right)\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (đk: t^o, H₂SO₄ đặc)

Xét tỉ lệ: \(\dfrac{1}{1}>\dfrac{\dfrac{15}{23}}{1}\), ta được CH₃COOH dư.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{C_2H_5OH}=\dfrac{15}{23}\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{15}{23}.90\%=\dfrac{27}{46}\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{27}{46}.88=\dfrac{1188}{23}\left(g\right)\)

dầu ăn nổi ở trên vì dầu ăn nhệ hơn nước và vớt dầu ăn ra còn lại là nước 

Dùng phương pháp chiết

- Acid: 

+ \(HCl\): hydrochloric acid

+ \(H_2SO_4\): sulfuric acid

+ \(CH_3COOH\): acetic acid

- Base:

+ \(Ca\left(OH\right)_2\): calcium hydroxide

+ \(KOH\): potassium hydroxide

- Oxide acid: \(SO_3\): sulfur trioxide

- Oxide base: \(FeO\): iron (II) oxide

- Muối: 

+ \(BaSO_4\): barium sulfate

+ \(NaCl\): sodium chloride