\(a)2Al + 6CH_3COOH \to 2(CH_3COO)_3Al + 3H_2\\ b)n_{Al} = \dfrac{2,7}{27} = 0,1(mol) ; n_{CH_3COOH} = \dfrac{200.10\%}{60} = \dfrac{1}{3}(mol)\\ n_{CH_3COOH} = \dfrac{1}{3}> 3n_{Al} = 0,3 \to CH_3COOH\ dư\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol) \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{CH_3COOH\ pư} = 3n_{Al} =0,3(mol) \Rightarrow m_{CH_3COOH\ pư} = 0,3.60 = 18(gam)\\ c) m_{dd} = 2,7 + 200 - 0,15.2 = 202,4(gam)\\ n_{(CH_3COO)_3Al} = n_{Al} = 0,1(mol)\\ m_{CH_3COOH\ dư} = 200.10\% - 18 = 2(gam)\\ C\%_{(CH_3COO)_3Al} = \dfrac{0,1.204}{202,4}.100\% = `10,08\%\\ \)

\(C\%_{CH_3COOH} = \dfrac{2}{202,4}.100\% = 0,988\%\)

$a\big)2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$b\big)$

$n_{Al}=\dfrac{2,7}{27}=0,1(mol)$

Theo PT: $n_{H_2SO_4}=1,5n_{Al}=0,15(mol)$

$\to m_{dd\,H_2SO_4}=\dfrac{0,15.98}{30\%}=49(g)$

$c\big)$

Theo PT: $n_{H_2}=0,15(mol);n_{Al_2(SO_4)_3}=0,05(mol)$

$\to V_{H_2}=0,15.22,4=3,36(l)$

$\to m_{Al_2(SO_4)_3}=0,05.342=17,1(g)$

\(m_{CH_3COOH}=150.12\%=18g\)

\(n_{CH_3COOH}=\dfrac{18}{60}=0,3mol\)

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

      0,3                   0,15                0,3                 0,15               ( mol )

\(m_{ddNa_2CO_3}=\left(0,15.106\right):10,6\%=150g\)

\(V_{CO_2}=0,15.22,4=3,36l\)

\(m_{CH_3COONa}=0,3.82=24,6g\)

\(m_{ddspứ}=150+150-0,15.44=293,4g\)

\(C\%_{CH_3COONa}=\dfrac{24,6}{293,4}.100=8,28\%\)

\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)

a) $Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$

b) $n_{Na_2CO_3} = \dfrac{21,2}{106} = 0,2(mol)$

$n_{HCl} =2 n_{Na_2CO_3} = 0,4(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,4}{0,4} = 1M$

c) $n_{CO_2} = n_{Na_2CO_3} = 0,2(mol) \Rightarrow V_{CO_2} = 0,2.22,4 = 4,48(lít)$

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2mol\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

 0,2               0,4           0,4        0,2        0,2

\(C_{M_{HCl}}=\dfrac{0,4}{0,4}=1M\)

\(V_{CO_2}=0,2\cdot22,4=4,48\left(l\right)\)

a) 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

b) \(n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

                     0,4------->0,2------------------------->0,2

=> \(m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)

=> \(m_{dd.Na_2CO_3}=\dfrac{21,2.100}{50}=42,4\left(g\right)\)

c) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)