2K + 2H2O ------> 2KOh + H2
2KOh + Cu(NO3)2 ------>2 KNO3 + Cu(OH)2
n H2 = 3,36/22.4=0.15 mol
=> n K = 0.3 mol
=> m K = 0.3*39=11.7 g
n KOh(tao thành) = 0.3 mol
n Cu(NO3)2=0.2*0.5=0.1 mol
=> sau phản ưng KOH dư
=> n Cu(OH)2=0.1 mol
=> m kết tủa= 0.1*98=9.8 g