Question

Answer 1

\(n_{CuO}=\dfrac{3.2}{80}=0.04\left(mol\right)\)

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

\(0.04.......0.04............0.04\)

\(m_{dd_{H_2SO_4}}=\dfrac{0.04\cdot98}{4.9\%}=80\left(g\right)\)

\(m_{dd}=3.2+80=83.2\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.04\cdot160}{83.2}\cdot100\%=7.69\%\)

Answer 2

a) PTHH: CuO+H₂SO₄ \(\rightarrow\)CuSo₄+H₂O

                0,04    0,04        0,04       0,04

Ta có n_CuO=\(\dfrac{3.2}{64+16}=0.04\) (mol)

b)Ta có m_H2SO4=\(0.04\cdot98=3.92\left(g\right)\)

             mdd_H2SO4=\(\dfrac{m_{H2SO4}\cdot100}{4,9}=\dfrac{3.92\cdot100}{4.9}=80\left(g\right)\)

Ta có mdd_CuSO4=m_CuO+mdd_H2SO4=3,2+80=83,2 (g)

         m_CuSO4=\(0.04\cdot160=6.4\) (g)

\(\Rightarrow C\%=\dfrac{6.4}{83.2}\cdot100\%=7.69\%\)

Tick nha bạn 😘