\(a.n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\right]=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)

1.

Theo đề bài ta có : \(\left\{{}\begin{matrix}nAl=\dfrac{0,54}{27}=0,02\left(mol\right)\\nH2SO4=\dfrac{120.4,9}{100.98}=0,06\left(mol\right)\end{matrix}\right.\)

PTHH :

\(2Al+3H2SO4->Al2\left(So4\right)3+3H2\uparrow\)

0,02mol...0,03mol.......0,01mol.............0,03mol

Theo PTHH ta có : nAl = \(\dfrac{0,02}{2}mol< nH2SO4=\dfrac{0,06}{2}mol=>nH2SO4\left(dư\right)\) ( tính theo nal)

=> VH2(đktc) = 0,03.22,4 = 6,72(l)

=> \(\left\{{}\begin{matrix}C\%ddH2SO4\left(dư\right)=\dfrac{\left(0,06-0,03\right).98}{0,54+120-0,03.2}.100\%\approx2,44\%\\C\%ddAl2\left(SO4\right)3=\dfrac{0,01.302}{0,54+120-0,03.2}.100\%\approx2,5\%\end{matrix}\right.\)

Theo đề bài ta có : nNa2O = \(\dfrac{15,5}{62}=0,25\left(mol\right)\)

a) PTHH :

\(Na2O+H2O->2NaOH\)

0,25mol....0,25mol.....0,5mol

b) Nồng độ mol dd A là :

CMddNaOH = 0,5/0,5 = 1(M)

c) PTHH :

\(2NaOH+H2SO4->Na2SO4+H2O\)

0,5mol.........0,25mol

=> mddH2SO4 = \(\dfrac{0,25.98}{20}.100=122,5\left(g\right)=>VddH2SO4=\dfrac{122,5}{1,14}\approx107,5\left(ml\right)\)