$m_{dd\rm spu}=169+490=659g$

$n_{KOH}=\dfrac{169.20\%}{56}=0,604(mol)$

$n_{H_2SO_4}=\dfrac{490.10\%}{98}=0,5(mol)$

$2KOH+H_2SO_4\to K_2SO_4+2H_2O$

$\dfrac{0,604}{2}<0,5\to H_2SO_4$ dư 

$n_{K_2SO_4}=\dfrac{n_{KOH}}{2}=0,302(mol)$

$\to C\%_{K_2SO_4}=\dfrac{0,302.174.100}{659}=7,97\%$

$n_{H_2SO_4\rm dư}=0,5-0,302=0,198(mol)$

$\to C\%_{H_2SO_4}=\dfrac{0,198.98.100}{659}=2,94\%$

$ m_{KOH} = 169 × 20\% = 33,8 (g) $ 

$ \to n_{KOH} = \dfrac{ m}{ M} = \dfrac{ 33,8 }{ 56 } = \dfrac{ 169 }{ 280 } (mol) $ 

$ m_{H_2SO_4} = 490 × 10\% = 49 (g) $ 

$ \to n_{H_2O_4} = \dfrac{m}{M}  = \dfrac{ 49 }{ 98 } = 0,5 (mol) $ 

Phương trình :
$ 2KOH + H_2SO_4 \to K_2SO_4 + 2H_2O $

Xét số mol giữa $KOH$ và $H_2SO_4$ : $ \dfrac{  \dfrac{ 169 }{ 280 } }{ 2 } < \dfrac{ 0,5 }{ 1 } $ 

$ - n_{H_2SO_4}$ dư tính theo $n_{KOH}$ 

Theo phương trình : 

$ n_{K_2SO_4} = \dfrac{1}{2} × n_{KOH} = \dfrac{ 1 }{ 2 } × \dfrac{ 169 }{ 280 } = \dfrac{169}{560} (mol) $ 

$ \to m_{K_2SO_4} = n × M = \dfrac{169}{560} × 174 = \dfrac{ 14703 }{ 280 } (g) $ 

$ m_{dd} = m_{ddKOH} + m_{ddH_2SO_4} = 169 + 490 = 659 (g) $ 

$ \to C\%_{K_2SO_4} = \dfrac{m_{ct}}{m_{dd}} × 100\% = \dfrac{ \dfrac{ 14703 }{ 280 } }{ 659 } × 100 = 7,968\% $ 

$ n_{H_2SO_4 ( dư ) } = 0,5 - \dfrac{169}{560} = \dfrac{111}{560} (mol) $ 

$ \to m_{H_2SO_4 ( dư ) }  = n × M = \dfrac{111}{560}  × 98 = 19,425 (g) $ 

$ \to C\%_{H_2SO_4 ( dư ) } = \dfrac{m_{ct}}{m_{dd}} × 100\% = \dfrac{ 19,425 }{ 659 } × 100 = 2,948 \% $