Đáp án:

$a\big)1M$

$b\big)\begin{cases} V_{dd\,H_2SO_4}\approx 107,46(ml) \\ C_{M\,muối}\approx 0,41M \end{cases}$

Giải thích các bước giải:

$a\big)$

$n_{Na_2O}=\frac{15,5}{62}=0,25(mol)$

$Na_2O+H_2O\to 2NaOH$

Theo PT: $n_{NaOH}=2n_{Na_2O}=0,5(mol)$

$\to C_{M\,NaOH}=\frac{0,5}{0,5}=1M$

$b\big)$

$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

Theo PT: $n_{H_2SO_4}=\frac{1}{2}n_{NaOH}=0,25(mol)$

$\to m_{dd\,H_2SO_4}=\frac{0,25.98}{20\%}=122,5(g)$

$\to V_{dd\,H_2SO_4}=\frac{122,5}{1,14}\approx 107,46(ml)$

Đổi $107,46ml=0,10746l$

Theo PT: $n_{Na_2SO_4}=n_{H_2SO_4}=0,25(mol)$

$\to C_{M\,Na_2SO_4}=\frac{0,25}{0,5+0,10746}\approx 0,41M$

`n_{Na_2O}=(15,5)/(62)=0,25(mol)`

`a,`

PTHH:

`Na_2O+H_2O->2NaOH`

`0,25`              `->0,5(mol)`

`C_{M NaOH}=(0,5)/(0,5)=1(M)`

`b,`

PTHH:

`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`

`0,5`         `->0,25`   `->0,25(mol)`

`=>m_{H_2SO_4}=0,25.98=24,5(g)`

`m_{dd H_2SO_4}=24,5:20\%=122,5(g)`

`V=122,5:1,14≈107,46(ml)`

`C_{M Na_2SO_4}=(0,25)/(107,46:1000+0,5)≈0,4(M)`