Đáp án:
a) 14,81%
b) 20g
c) 1,6g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
n{H_2}S{O_4} = \dfrac{{20 \times 9,8\% }}{{98}} = 0,02\,mol\\
= > nCuS{O_4} = nCuO = n{H_2}S{O_4} = 0,02\,mol\\
m{\rm{dd}}spu = 0,02 \times 80 + 20 = 21,6g\\
C\% CuS{O_4} = \dfrac{{0,02 \times 160}}{{21,6}} \times 100\% = 14,81\% \\
b)\\
CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
nNaOH = 2nCuS{O_4} = 0,04\,mol\\
m{\rm{dd}}NaOH = \dfrac{{0,04 \times 40}}{{8\% }} = 20g\\
c)\\
Cu{(OH)_2} \to CuO + {H_2}O\\
nCuO = nCu{(OH)_2} = 0,02\,mol\\
mCuO = 0,02 \times 80 = 1,6g
\end{array}\)
