Расставьте коэффициенты и определите тип реакции: а) hgcl2 + al = alcl3 + hg б) сa + p = ca3p2 в) h2o2 = h2o + o2 г) mgo + h3po4 = mg3 (po4)2 + h2o

1. 3HgCl₂ + 2Al → 2AlCl₃ + 3Hg

замещение

2. 3Ca + 2P → Ca₃P₂ соединение

3. 2H₂O₂ → 2H₂O + O₂ разложение

4. 3MgO + 2H₃PO₄ → Mg₃(PO₄)₂ + 3H₂O обмен

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Объяснение:

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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%