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(a) Urea is prepared by the reaction between ammonia and carbon dioxide. 2NH_(y(s))+CO_(2(g))to(NH_(2))_(2)CO_((aq))+H_(2)O_((2)) In one process, 637.2 g of NH, are allowed to react with 1142 g of CO_(2) (i)Which of the two reactants is the limiting reagent? (ii) Calculate the mass of (NH_(2))_(2) CO formed (im) How much of the excess reagent in grams is left at the end of the reaction?

Answer» <html><body><p></p>Solution :(a) (1) `underset(2 "moles")(2NH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>(g))+underset(1" mole "CO_(2)(g))tounderset(1" mole") ((NH_(4))_(2)CO_((aq)))+H_(2)O_(1)`<br/>No. of moles of ammonia -=`(637.2)/(17)`=37.45 mol e<br/>No. of moles of `CO_(2)=(1142)/(44)` = 25.95 moles <br/>As per the balanced equation, one mole of Co, requires 2 moles of ammonia. No. of moles of NH, required to <a href="https://interviewquestions.tuteehub.com/tag/react-613674" style="font-weight:bold;" target="_blank" title="Click to know more about REACT">REACT</a> with 25.95 moles of `CO_(2)`, is =`(2)/(1)xx25.95`=51.90moles<br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>`37.45 moles of NH is not enough to completely react with `CO_(2)` (25.95 moles). Hence, NH must be the limiting reagent, and`CO_(2)`, is excess reagent. <br/>(ii) 2 moles of ammonia produce mole of urea. `therefore` Limiting reagent 37.45 moles of `NH_(3)`, can produce `(1)/(2)xx37.45` moles of urea. = 18.725 moles of <a href="https://interviewquestions.tuteehub.com/tag/urca-3250618" style="font-weight:bold;" target="_blank" title="Click to know more about URCA">URCA</a>.<br/>The mass of 18.725 moles of urca =No. of moles` xx`Molar mass<br/>= `18.725xx60` =1123.5 g of urea.<br/>2 moles of ammonia requires mole of `CO_(2)` <br/>`therefore`Limiting reagent 37.45 mol es of NII, will require `(1)/(2)xx37.45mol es of CO_(2)xx37.45` moles of `CO_(2)` =18.725 moles of `CO_(2)` <br/>No. of moles of the excess reagent (CO) lell -2595-18725 - =7225<br/>The mass of the excess reagent `(CO_(2))`left =`7225xx44` = 3179g of `CO_(2)`</body></html>


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