Question

For the following reactions, find the following: a) Which of the reagents is the limiting reagent? b) What is the maximum amount of each product that can be formed? How much of the other reagent is left over after the reaction is complete? Consider the following reaction: 3 NH.NO3 + Na3PO4 → (NH4)3PO4 + 3 NaNO3 Answer the questions above, assuming we started with 30 grams of ammonium nitrate and 50 grams of sodium phosphate. 2) Consider the following reaction: 3 Caco, + 2 FePO4 → Cas(PO4)2 + Fe2(CO3)3 Answer the questions at the top of this sheet, assuming we start with 100 grams of calcium carbonate and 45 grams of iron (III) phosphate.

Answer 1

1. 3NHYNO₃ + Na₂ Poy - (NH 4 ) ₂ Poy + 3NaNO₃ Moles of NH4NO3 - Mass Molar Mass = 30g = 0.375 ml 8ogmatt Moles of Na, Poy = 50g = 0.305 mel. - 164gmely (a) We have to find limiting reagent. imol of Naz Poy reacts with 3 meles of NHYNO 0.305 mel Na₂ pou reacts with: 3 mel WHY NO3 x 0.305 mel Na Poy Imel No Pou = 0.915 med N HY NO. But there is only 0.75 mel of NHY NO Therefore NH y NO₂ will be the limiting Treagent . It will decide the amount of product formation 6.) 3 meles of N Hy No, forms I mes of (NH4)₂ Poy 0.375 met will produce : Imel (NH4)₂ Poy x 0.375 mol WHY NO₂ 3 mel NHy No = 0.125 mel (NH4)₂ Poy Mass of (NH4), Pon = 0.125 mel x 14g glmel = 18.625 g (NH4) apoy

Amount of NQNOZ: 3 meles of Nhy No produces 3 moles of Nano 10.375 mel will produce 3 me Nano X 0.375 mol NHANO 3 mol NHYNO, = 0.375 mel Nano mass of Nanoz = 0.375 ml x 85 glmel. Amount of NaNO3 = 31.875g. 1 = 0. 375 mol NaNO (c) Na poy present in excess. Number of Excess moles (Total moles) -Moles which is consumed.). 3 meles NHY Noz reacts with 1 mol of Na, Poy. 10.375 mer reacts with : Imel Nazpou x 0.375 NHO NA 3 me NHY NO₂ = 0.125 mel Na, Poy. Excess meres = 10.305 -0.125) meles of Napoy.. = 0.180 meles Naz Poy. left over amount = 0.180 mel x 164 g mel. = 29.5 9.