Đáp án:
a) 16g
b) 100g
c) 14,81%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{8}{{80}} = 0,1\,mol\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,1\,mol\\
{m_{CuS{O_4}}} = n \times M = 0,1 \times 160 = 16g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{CuO}} = 0,1\,mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,1 \times 98 = 9,8g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{m \times 100}}{{{C_\% }}} = \dfrac{{9,8 \times 100}}{{9,8}} = 100g\\
c)\\
{m_{{\rm{dd}}spu}} = 100 + 8 = 108g\\
{C_\% }CuS{O_4} = \dfrac{{16}}{{108}} \times 100\% = 14,81\%
\end{array}\)