Đáp án:
11,26g
Giải thích các bước giải:
\(\begin{array}{l}
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
nCuO = \dfrac{8}{{80}} = 0,1\,mol\\
n{H_2}S{O_4} = nCuO = 0,1\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,1 \times 98}}{{9,8\% }} = 100g\\
m{H_2}O(bd) = 100 - 0,1 \times 98 = 90,2g\\
m{H_2}O\,spu = 90,2 + 0,1 \times 18 = 92g\\
mCuS{O_4}.5{H_2}O = a(g)\\
mCuS{O_4}\,kt = 0,64a(g)\\
mCuS{O_4}\,bd = 0,1 \times 160 = 16g\\
mCuS{O_4} = 16 - 0,64a(g)\\
m{H_2}Okt = a - 0,64a = 0,36a(g)\\
m{H_2}O = 92 - 0,36a(g)\\
\dfrac{{16 - 0,64a}}{{92 - 0,36a}} = \dfrac{{10}}{{100}} \Rightarrow a = 11,26g
\end{array}\)