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Уравнение реакции - 2 So2 + O2 = 2 SO3 44.8 литров оксида серы это 44.8 \ 22.4 = 2 моль следовательно мы должны получить 2 моль оксида серы 3 ( т.к по уравнению отношение моль оксида серы 2 и оксида серы 3 - 1:1) далее считаем теоретическую массу по данному уравнению ( а практическая - то, что мы получили (масса, данная в условии)) 2 моль оксида серы (3 ) это 2 *80 = 160 грамм теперь найдем выход реакции по формуле ( M практическое \ М теоретическое * 100% ) это будет 132 \ 160 * 100% = 0.825 * 100 = 82.5% ответ: выход реакции 82.5%
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Объяснение:
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44.8 литров оксида серы это 44.8 \ 22.4 = 2 моль
следовательно мы должны получить 2 моль оксида серы 3 ( т.к по уравнению отношение моль оксида серы 2 и оксида серы 3 - 1:1)
далее считаем теоретическую массу по данному уравнению ( а практическая - то, что мы получили (масса, данная в условии))
2 моль оксида серы (3 ) это 2 *80 = 160 грамм
теперь найдем выход реакции по формуле ( M практическое \ М теоретическое * 100% ) это будет 132 \ 160 * 100% = 0.825 * 100 = 82.5%
ответ: выход реакции 82.5%