Đáp án:
a) 10g
b) 2,45g
c) 6,17%
d) 2g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
nCuS{O_4} = \dfrac{{50 \times 8\% }}{{160}} = 0,025\,mol\\
nNaOH = 2nCuS{O_4} = 0,05\,mol\\
m{\rm{dd}}NaOH = \dfrac{{0,05 \times 40}}{{20\% }} = 10g\\
b)\\
nCu{(OH)_2} = nCuS{O_4} = 0,025\,mol\\
\Rightarrow mCu{(OH)_2} = 0,025 \times 98 = 2,45\,g\\
c)\\
m{\rm{dd}}spu = 50 + 10 - 2,45 = 57,55g\\
C\% N{a_2}S{O_4} = \dfrac{{0,025 \times 142}}{{57,55}} \times 100\% = 6,17\% \\
d)\\
Cu{(OH)_2} \to CuO + {H_2}O\\
nCuO = nCu{(OH)_2} = 0,025\,mol\\
mCuO = 0,025 \times 80 = 2g
\end{array}\)