7)

5 tấn=5000 kg

Để xử lý 5 tấn hạt giống cần thể tích dung dịch \(CuSO_4\) là

\({V_{dd{\text{ CuS}}{{\text{O}}_4}}} = \frac{{5000}}{{100}}.8 = 400{\text{ lít = 400000 ml}}\)

\( \to {m_{dd{\text{CuS}}{{\text{O}}_4}}} = 400000.1 = 400000{\text{ gam }}\)

\( \to {m_{CuS{O_4}}} = 400000.0,02\%  = 80{\text{ gam}}\)

\( \to {n_{CuS{O_4}}} = \frac{{80}}{{64 + 32 + 16.4}} = 0,5{\text{ mol = }}{{\text{n}}_{CuS{O_4}.5{H_2}O}}\)

\( \to {m_{CuS{O_4}.5{H_2}O}} = 0,5.(64 + 32 + 16.4 + 18.5) = 125{\text{ gam}}\)

8)

Ta có:

\({n_{{{(N{H_4})}_2}S{O_4}}} = \frac{{200}}{{18.2 + 96}} = 1,515{\text{ kmol}} \to {{\text{n}}_N} = 2{n_{{{(N{H_4})}_2}S{O_4}}} = 3,03{\text{ kmol}}\)

\( \to {m_N} = 3,03.14 = 42,42{\text{ kg}}\)

\( \to \% {m_N} = \frac{{42,42}}{{15.1000}}.100\%  = 0,2828\% \)

\({n_{Ca{{({H_2}P{O_4})}_2}}} = \frac{{50}}{{40 + (1.2 + 31 + 16.4).2}} = 0,2137{\text{ kmol = }}{{\text{n}}_{{P_2}{O_5}}}\)

\( \to {m_{{P_2}{O_5}}} = 0,2137.142 = 30,3454{\text{ kg}}\)

\( \to \% {m_{{P_2}{O_5}}} = \frac{{30,3454}}{{15000}} = 0,2023\% \)

Ta có:

\({n_{KCl}} = \frac{{30}}{{39 + 35,5}} = 0,4027{\text{ kmol}} \to {{\text{n}}_{{K_2}O}} = 0,20135{\text{ kmol}}\)

\( \to {m_{{K_2}O}} = 0,20135.(39.2 + 16) = 18,9269{\text{ kg}}\)

\( \to \% {m_{{K_2}O}} = \frac{{18,9269}}{{15000}} = 0,12618\% \)