a)nKOH = \(\dfrac{11,2.140}{100.56}\) = 0,28 mol
2KOH + CuSO4 -> Cu(OH)2 \(\downarrow\)+ K2SO4
0,28---->0,14----->0,14-------->0,14
b) ta có :4%=\(\dfrac{0,14.160}{140+m}.100\%\)
=>m = 420g
d)
C%K2SO4 = \(\dfrac{0,14.174}{140+420-0,14.98}.100\%\)
= 4,45%