a)
Ta có:
\({n_{{{(N{H_4})}_2}S{O_4}}} = \frac{{120}}{{(14 + 1.4).2 + 32 + 16.4}} = \frac{{10}}{{11}}\)
\( \to {n_N} = 2{n_{{{(N{H_4})}_2}S{O_4}}} = \frac{{20}}{{11}}\)
\( \to {m_N} = \frac{{20}}{{11}}.14 = \frac{{280}}{{11}}{\text{ gam}}\)
\( \to \% {m_N} = \frac{{\frac{{280}}{{11}}}}{{120}} = 21,21\% \)
b)
Ta có:
\({n_{CaHP{O_4}}} = \frac{{80}}{{40 + 1 + 31 + 16.4}} = \frac{{10}}{{17}}{\text{ mol = }}{{\text{n}}_P}\)
\( \to {n_{{P_2}{O_5}}} = \frac{1}{2}{n_P} = \frac{5}{{17}}{\text{ mol}}\)
\( \to {m_{{P_2}{O_5}}} = \frac{5}{{17}}.(31.2 + 16.5) = \frac{{710}}{{17}}{\text{ gam}}\)
\( \to \% {m_{{P_2}{O_5}}} = \frac{{\frac{{710}}{{17}}}}{{80}} = 52,2\% \)
c)
Ta có:
\({n_{KCl}} = \frac{{60}}{{39 + 35,5}} = \frac{{120}}{{149}}{\text{ mol = }}{{\text{n}}_K}\)
\( \to {n_{{K_2}O}} = \frac{1}{2}{n_K} = \frac{{60}}{{149}}{\text{ mol}}\)
\( \to {m_{{K_2}O}} = \frac{{60}}{{149}}.(39.2 + 16) = \frac{{5640}}{{149}}{\text{ gam}}\)
\( \to \% {m_{{K_2}O}} = \frac{{\frac{{5640}}{{149}}}}{{60}} = 63,09\% \)
