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Trả lời:
$a, Pt1: CuSO_4+2NaOH \to Cu(OH)_2+Na_2SO_4$
$Pt2: Cu(OH)_2 \xrightarrow{t^o} CuO+H_2O$
$b, n_{CuSO_4}=0,5.0,4=0,2(mol)$
$⇒n_{NaOH}=2.n_{CuSO_4}=2.0,2=0,4(mol)$
$⇒m_{NaOH}=0,4.40=16(g)$
$⇒m_{ddNaOH}=\dfrac{16.100}{20}=80(g)$
$c, n_{CuO}=n_{Cu(OH)_2}=n_{CuSO_4}=0,2(mol)$
$⇒m_{CuO}=0,2.80=16(g).$