$n_{KOH}=\dfrac{200.7\%}{56}=0,25(mol)$
$n_{ZnSO_4}=\dfrac{100.16,1\%}{161}=0,1(mol)$
$ZnSO_4+2KOH\to Zn(OH)_2+K_2SO_4$
$\Rightarrow KOH$ dư
$n_{Zn(OH)_2}=0,1(mol)$
$n_{K_2SO_4}=0,1.2=0,2(mol)$
$n_{KOH\text{dư}}=0,25-0,1.2=0,05(mol)$
$Zn(OH)_2+2KOH\to K_2ZnO_2+2H_2O$
$\Rightarrow Zn(OH)_2$ dư
$n_{K_2ZnO_2}=n_{Zn(OH)_2\text{pứ}}=0,025(mol)$
$\Rightarrow n_{Zn(OH)_2\text{dư}}=0,1-0,025=0,075(mol)$
$\Rightarrow m_{dd\text{spu}}=200+100-0,075.99=292,575g$
$C\%_{K_2SO_4}=\dfrac{0,2.174.100}{292,575}=11,89\%$
$C\%_{K_2ZnO_2}=\dfrac{0,025.175.100}{292,575}=1,5\%$
