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Đáp án:
7.
\(M_{hh}^{} = 16,75.2 = 33,5{\text{ g/mol}}\)
\(\frac{{n_{NO}^{}}}{{n_{N_2^{}O}^{}}} = \frac{{44 - 33,5}}{{33,5 - 30}} = \frac{3}{1}\)
\(n_{NO}^{}:n_{N_2^{}O}^{} = 3x:x\)
\(Al \to A{l^{3 + }} + 3e\)
\({\text{ 0}}{\text{,02 0}}{\text{,06}}\)
\(Fe \to F{e^{3 + }} + 3e\)
\(\;{\text{0}}{\text{,01 0}}{\text{,03}}\)
\({N^{ + 5}} + 3e \to {N^{ + 2}}\)
\({\text{ 9x 3x}}\)
\(2{N^{ + 5}} + 8x \to N_2^{ + 1}\)
\({\text{ 8x x}}\)
\(BTelectron:9x + 8x = 0,09\)
\( \to x = \frac{9}{{1700}}\;{\text{mol}}\)
\(V = (3x + x).22,4 = 0,474l\)
\(8.\)
\(M_{hh}^{} = 14,75.2 = 29,5{\text{ g/mol}}\)
\(\frac{{n_{NO}^{}}}{{n_{N_2^{}}^{}}} = \frac{{29,5 - 28}}{{30 - 29,5}} = \frac{3}{1}\)
\(n_{NO}^{}:n_{N_2^{}}^{} = 3x:x\)
\(Al \to A{l^{3 + }} + 3e\)
\({\text{ 0}}{\text{,05 0}}{\text{,15}}\)
\(Fe \to F{e^{3 + }} + 3e\)
\(\;{\text{0}}{\text{,01 0}}{\text{,03}}\)
\(Cu \to C{u^{2 + }} + 2e\)
\({\text{ 0}}{\text{,05 0}}{\text{,1}}\)
\({N^{ + 5}} + 3e \to {N^{ + 2}}\)
\({\text{ 9x 3x}}\)
\(2{N^{ + 5}} + 10x \to N_2^{ + 1}\)
\({\text{ 10x x}}\)
\( \to {\text{9x + 10x = 0}}{\text{,15 + 0}}{\text{,1 + 0}}{\text{,03}}\)
\( \to x = \frac{7}{{475}}\;{\text{mol}}\)
\(V = (3x + x).22,4 = 1,32l\)
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