Question

300.0 g of NH3 reacts with 800.0 g of sulfuric acid, to produce 900.0 g of ammonium sulfate according to the chemical equation below. Which of the two reactants would be the limiting reactant? Show calculations What is the theoretical yield of ammonium sulfate? What is the percent yield? How much excess reactant is left behind?

Answer 1

Se quen reaction can be written as - NH + H2SO WHISO The balanced chemical equation L 2NH + H2SO (Nta), so Jo, from the balanced equation, З тал ми е м І тя and forms t more of (NH4)2SO . 5904- a moks of Nty contains 2x ( 14.006 + 3×1,008) of NA - _ =2217.03 9 of Naty = 34.06 g of Nithy I mole of so contains (2x1,008 + 32.06 +4415.99) 9. - 98-039 of 184 I mble of (nula), son containgel 14.00 +4x 1.002) +33,06 +4x15.09 -2 (18.03 ) + 96.029 - 132.06 9 of Wola), 2 mils of the react with Imous of 450a. 34.06g of Niy reacts with 98.039 of 4 soa Gwew that 300 g of the reacts with soo gofyse. 300 g of Nihy wil react with 98-03 300 g of the 34.06x song of H50g = 863,44 g of HS But only 800 g of 50g is grew. So 450 is the limiting reagent as 863.49 a is required to react with 300 e g of hos produd will depend on by soa. ed to react with 300 g of Ni. To the

As calculated above an Imole of H₂50g reacts with 2 mole of with and produced I mole of (NHa), sog Lic 98.039 of isoa produced 132,08 g of (NH4)₂SO4 1.800 g of 1280 will produces 132,08 x 800 of (N14250 a = 1077.87 g of NHa), Sog Theoretical yuld of (NH4) So is 1077, 879. yield Percentage yield = experimental Theoretical yield 900 X 10% [ 1077.87 a 83.49% 900g of NH Soa. is given] Percentage yuld = 83.49% As calculated above, con 98.039 of 150g reacts with 34.06 g of NH3

i 30 g of 14.50 will react with 34006 800 g Ny 98.03 < = 277,95 9 of Nity Only 277.95 g of Nithy is used to react with 800 of of thon so remaining is present as excen as 300 g of Nithy is given anun reactant left behind = 1300-277.95) 22.059 22,05 g of Nt is left behin