Question

Answer 1

Answer 2

\(n_{CuSO_4}=1.0,2=0,2(mol);n_{NaOH}=1.0,3=0,3(mol)\\ PTHH:CuSO_4+2NaOH\to Cu(OH)_2\downarrow+Na_2SO_4\)

Vì \(\dfrac{n_{NaOH}}{2}<\dfrac{n_{CuSO_4}}{1}\) nên \(CuSO_4\) dư

\(a,n_{Cu(OH)_2}=\dfrac{1}{2}n_{NaOH}=0,15(mol)\\ \Rightarrow a=m_{Cu(OH)_2}=0,15.98=14,7(g)\\ b,n_{CuSO_4(dư)}=0,2-0,15=0,05(mol)\\ n_{Na_2SO_4}=0,15(mol)\\ \Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,15}{0,2+0,3}=0,3M\\ C_{M_{Cu(OH)_2}}=\dfrac{0,05}{0,2+0,3}=0,1M\\ c,PTHH:Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=n_{Cu(OH)_2}=0,15(mol)\\ \Rightarrow b=m_{CuO}=0,15.80=12(g)\)