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Urea is prepared by the reaction between ammonia and carbon dioxide.2NH_(3_((g)))+CO_(2_((g)))rarr(NH_(4))_(2)CO_(aq)+H_(2)O_((l))In one process, 637.2 g of NH_(3)are allowed to react with 1142 g of CO_(2)(a) Which of the two reactants is the limiting reagent? 2NH_(3_((g)))+CO_(2_((g)))rarr(NH_(4))_(2)CO_(aq)+H_(2)O_((l))(b) Calculate the mass of (NH_(4))_(2)CO formed. 2NH_(3_((g)))+CO_(2_((g)))rarr(NH_(4))_(2)CO_(aq)+H_(2)O_((l))(c) How much of the excess reagent in grams is left at the end of the reaction? |
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Answer» Solution :(a) `2NH_(3_((g))) + CO_(2((g))) rarr (NH_(4))_(2)CO_(aq) + H_(2)O_((l))` No. of moles of ammonia =`637.3/17` = 37.45 MOLE No. of moles of `CO_(2)` =`1142/44`=25.95 moles As per the balanced equation, one mole of `CO_(2)` requires 2 moles of ammonia. `:.` No. of moles of `NH_(3)` REQUIRED to react with 25.95 moles of `CO_(2)` is = `2/1 xx 25.95` = 51.90 moles. `:.` 37.45 moles of `NH_(3)` is not ENOUGH to completely react with `CO_(2)` (25.95 moles). HENCE, `NH_(3)` must be the limiting reagent, and `CO_(2)` is excess reagent. (b) 2 moles of ammonia produce 1 mole of urea. `:.` Limiting reagent 37.45 moles of `NH_(3)` can produce `1/2 xx 37.45`moles of urea = 18.725 moles of urea. The mass of 18.725 moles of urea = No. of moles `xx` Molar mass = 18.725 `xx` 60 = 1123.5 g of urea. (C) 2 moles of ammonia requires 1 mole of `CO_(2)` `:.` Limiting reagent 37.45 moles of `NH_(3)` will require` 1/2 xx 37.45` moles of `CO_(2)` = 18.725 moles of `CO_(2)` `:.` No. of moles of the excess reagent `(CO_(2))` left = 25.95- 18.725 — 7.225 The mass of the excess reagent `(CO_(2))` left = `7.225xx44` = 317.9 g of `CO_(2)` . |
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