Đáp án:

$a)\ m_{NaOH}=8(g)$

$b)\ V_{KOH}=38,28(ml)$

Giải thích các bước giải:

$a)$

$n_{H_2SO_4}=0,02×1=0,02(mol)$

$H_2SO_4+2NaOH\to Na_2SO_4+2H_2O$ 

$n_{NaOH}=2n_{H_2SO_4}=0,04(mol)$

$m_{NaOH}=0,04×40=1,6(g)$

$m_{dd\ NaOH}=\dfrac{1,6}{20\%}=8(g)$

$b)$

$H_2SO_4+2KOH\to K_2SO_4+2H_2O$ 

$n_{KOH}=2n_{H_2SO_4}=0,04(mol)$

$m_{KOH}=0,04×56=2,24(g)$

$m_{dd\ KOH}=\dfrac{2,24}{5,6\%}=40(g)$

$V_{KOH}=\dfrac{40}{1,045}=38,28(ml)$

`#Ly`

$a)$

Đổi $20$ $ml$ $=$ $0,02$ $l$

`n_(H_2SO_4` `=` `0,02.1=0,02` $(mol)$

PTHH: $H_2SO_4$ $+$ $2NaOH$ `->` $Na_2SO_4$ $+$ $2H_2O$

Theo PT, `n_(NaOH` `=` `2n_(H_2SO_4` `=` `0,04` $(mol)$

`m_(NaOH` `=` `0,04.40=1,6` $(g)$

`m_(ddNaOH` `=` `(1,6.100)/(20)=8` $(g)$

$b)$

PTHH: $H_2SO_4$ $+$ $2KOH$ `->` $K_2SO_4$ $+$ $2H_2O$

Theo PT, `n_(KOH` `=` `2n_(H_2SO_4` `=` `0,04` $(mol)$

`m_(KOH` `=` `0,04.56=2,24` $(g)$

`m_(ddKOH` `=` `(22,4.100)/(5,6)=40` $(g)$

`V_(ddKOH` `=` `(40)/(1,045)=38,2775` $(ml)$