Question

Answer 1

n_CuO=\(\dfrac{16}{80}=0,2mol\)

CuO+H₂SO₄->CuSO₄+H₂O

0,2----0,2------------0,2-----0,2

->C%=\(\dfrac{0,2.98}{200}100\)=9,8%

b)CuSO₄+2NaOH->Cu(OH)₂+Na₂SO₄

0,2---------------0,4--------0,2---------------0,2

=>a=m_Cu(OH)2=0,2.98=19,6g

=>CM _H2SO4=\(\dfrac{0,2}{0,4}=0,5M\)

Answer 2

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

 0,2         0,2               0,2                 ( mol )

\(C\%_{H_2SO_4}=\dfrac{0,2.98}{200}.100=9,8\%\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

 0,2              0,4                0,2                    ( mol )

\(a=m_{Cu\left(OH\right)_2}=0,2.98=19,6\left(g\right)\)

\(C_{M_{NaOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)