a)
\(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PTHH: Cu(OH)2 --to--> CuO + H2O
______0,3<-------------0,3
=> mCu(OH)2 = 0,3.98 = 29,4 (g)
b)
PTHH: CuO + 2HNO3 --> Cu(NO3)2 + H2O
______0,3--->0,6---------->0,3
=> \(C_{M\left(HNO_3\right)}=\dfrac{0,6}{0,2}=3M\)
mCu(NO3)2 = 0,3.188 = 56,4 (g)