a,
$3CH_3COOH+Al\to (CH_3COO)_3Al+1,5H_2$
b,
$n_{Al}=\dfrac{2,7}{27}=0,1(mol)$
$n_{CH_3COOH}=\dfrac{200.10\%}{60}=0,333(mol)$
$\Rightarrow CH_3COOH$ dư
$n_{H_2}=1,5n_{Al}=0,15(mol)$
$\Rightarrow V_{H_2}=0,15.22,4=3,36l$
$n_{CH_3COOH\text{pứ}}=3n_{Al}=0,3(mol)$
$\Rightarrow m_{CH_3COOH\text{pứ}}=18g$
c,
$m_{dd\text{sau}}=2,7+200-0,15.2=202,4g$
$\Rightarrow C\%_{(CH_3COO)_3Al}=\dfrac{0,1.204.100}{202,4}=10,08\%$
$n_{CH_3COOH\text{du}}=0,333-0,3=0,033(mol)$
$\Rightarrow C\%_{CH_3COOH}=\dfrac{0,033.60.100}{202,4}=1\%$