a) \(n_{CH_3COOH}=0,1.0,3=0,03\left(mol\right)\)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                  0,03---->0,03--------->0,03

=> \(V_{dd.NaOH}=\dfrac{0,03}{1,5}=0,02\left(l\right)\)

b) m_CH3COONa = 0,03.82 = 2,46 (g)

c) \(C_{M\left(CH_3COONa\right)}=\dfrac{0,03}{0,1+0,02}=0,25M\)

a) \(n_{Al}=\dfrac{32,4}{27}=1,2\left(mol\right)\)

PTHH: 2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu

_____1,2--->1,8-------->1,2----->1,8

=> m_Cu = 1,8.64 = 115,2 (g)

b) \(V_{ddCuCl_2}=\dfrac{1,8}{1,5}=1,2\left(l\right)\)

c) \(AlCl_3+3NaOH\rightarrow3NaCl+Al\left(OH\right)_3\downarrow\)

\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

a, \(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Fe}=1,5\left(mol\right)\Rightarrow V_{H_2}=1,5.22,4=33,6\left(l\right)\)

b, \(n_{HCl}=2n_{Fe}=3\left(mol\right)\Rightarrow m_{HCl}=3.36,5=109,5\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{109,5}{10,95\%}=1000\left(g\right)\)

c, \(n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\)

Ta có: m _{dd sau pư} = 84 + 1000 - 1,5.2 = 1081 (g)

\(\Rightarrow C\%_{FeCl_2}=\dfrac{1,5.127}{1081}.100\%\approx17,62\%\)

\(n_{H_2SO_4}=0,1.3=0,3\left(mol\right)\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo PT: \(n_{Zn}=n_{H_2SO_4}=0,3\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(n_{AgNO_3}=0,1.0,2=0,02\left(mol\right)\\ n_{HCl}=0,3.0,2=0,06\left(mol\right)\\ AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ a,Vì:\dfrac{0,02}{1}< \dfrac{0,06}{1}\Rightarrow HCldư\\ \Rightarrow n_{AgCl}=n_{HNO_3}=n_{AgNO_3}=0,02\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,02=2,87\left(g\right)\\ b,dd.sau.p.ứ:HNO_3,HCl\left(dư\right)\\ n_{HCl\left(dư\right)}=0,06-0,02=0,04\left(mol\right)\\V_{ddsau}=V_{ddAgNO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ \Rightarrow C_{MddHCl\left(dư\right)}=\dfrac{0,04}{0,4}=0,1\left(M\right)\\ C_{MddHNO_3}=\dfrac{0,02}{0,4}=0,05\left(M\right)\)

\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

0,095          0,19              0,095            0,095

\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)

0,095                 0,095

\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)

a) \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: `Fe + 2HCl -> FeCl_2 + H_2`

           0,05->0,1----->0,05---->0,05

`=> V_{ddHCl} = (0,1)/2 = 0,05 (l)`

b) `V_{H_2} = 0,05.22,4 = 1,12 (l)`

c) `C_{M(FeCl_2)} = (0,05)/(0,05) = 1M`

a)       2Al + 3H₂SO₄  →  Al₂(SO₄)₃   +  3H₂

nH₂SO₄ = 0,2.1,5= 0,3 mol

nAl = \(\dfrac{5,4}{27}\) = 0,2 mol

Ta có \(\dfrac{nAl}{2}\)= \(\dfrac{nH_2SO_4}{3}\) => phản ứng vừa đủ , Al và H₂SO₄ đều hết. 

nH₂ = nH₂SO₄ = 0,3 mol => V H₂ = 0,3.22,4 = 6,72 lít

b) vì Al và H₂SO₄ đều phản ứng hết => dung dịch sau phản ứng chỉ gồm Al₂(SO₄)₃  0,2 mol

=> [Al₂(SO₄)₃] = \(\dfrac{0,2}{0,2}\) = 1M