Question

For the following reactions, find the following: Which of the reagents is the limiting reagent? 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 NH4NO3 + Na3PO4 → (NH4)3PO4 + 3 NaNO, 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 CaCO3 + 2 FePO4 → Cas(PO )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

91: The reaction is 3 NH4NO2 + Naz po4 —(NH4), PO4 + 3 NaNO3 From the chemical reaction, it is cleared that wehen 3 mole of NHG NO3 reacts with 1 mole of Na3PO4 then 1 mole of (NH), PO4 and 3 mole of Nanoz is formed. Molecular weight of NH4NO3 = 80 g mot Here 30 g of NHINO2 is used, so number of moles of NHANU3 used for reaction.is ..398 0.375 mol of NH 4 NO3 number of moley = - 3gg mät = 0.37 in the reaction , so, 50 g of Na, pou is used Number of moles of Naz P04=2 = 0.305 mol 163094gmol Molecular weight of Naz pou=163.94 gmoil

. From the balunud elemical equation . 1 mole of Naz piou requires = 3 mole of NH4NO3 ..0.305 mole of Naz Pou requines = 3x0.305 mole of NHCMO =0.915 mole of NH4NO3 However there are only 0.375 mole of NHqNUz is po avail for the reaction. So NH4NO3 is the limiting reagent font given chemical reaction. @ NHINO3 is the limiting reagent. 1 the

From the chemical equation it is seen that 3 mole of NHaNO3 enn form manimum of 3 mole NaNO3 I mole of NHANU, Can form manümum of 1 mole Nanoz 0.375 mole of NHINO, ca boom manimum of 0.375 mole Nanoz Again - 1 mole of Nazpou can form manimum of a mole of NH4), P04 0.305 mole of Naz pou can form marinum of 0.305 mole of (pH) po - So I mole af NaNO3 = 84.99 y moi 0.375 mole of Nanoz = 31.99 Aguin 1 mole of (NH4)₂PO4 = 132.06gmoil 0.305 mole of (NH4), PO4 = 409 Manimum amount of (NH4)₂PO4 = 40 g is formed Manimum amount of Nanoz = 31g is formed.

o As NHINO2 is the limiting reagent so reaction mainly depends on the amount of NHaNO3. from chemical equation 3 mole of NHINO3 required = 1 mole of Na₃PO4 0.375 mole of NHINO required = 0.375 X 1/2 mole Nazpo = 0.125 mole Nazpon. Howeven we have 0.305 mole of Naapoy.ie (0:305-0.125) mol o exy Naz Po: 10.305-0-125) mole = 0.18 mole of excess Na₃PO4 A mole Nazpou = 163.94g. 1.0.18 mole Naz pou - 163.949 x 0.18 = 29.50 g

@ Limiting reagent is NHE NO 6 Manimum amount of Naz pou can form & 40g Manimum amount of Nanoz can form 31 g € 29.5 g of Naz pou will left after completation of the reaction. VO