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Comprehension # 7 Oleum is considered as a solution of SO_(3) in H_(2)SO_(4), which is obtained by passing SO_(3) in solution of H_(2)SO_(4). When 100g sample of oleum is diluted with desired weight of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109gtotal mass of pure H_(2)SO_(4) will be formed when 100g of oleum is diluted by 9g of H_(2)O which combines combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)OrarrH_(2)SO_(4) If excess water is added into a bottle sample labelled as 112% H_(2)SO_(4) and is reacted with 5.3 g Na_(2)CO_(3) then find the volume of CO_(2) evlovedat 1atm pressure and 300 K temperature after the completion of the reaction |
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Answer» `2.46 L` `{:((112)/(98),(5.3)/(106),""0.05),(1.1428,0.05,):}` `V_(co^(2)) = 0.05xx24.63 = 1.23l` |
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