Question

For each of the following unbalanced chemical equations, suppose 1.20 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed). a. UO2 (s) + HF(aq) → UF4(aq) + H2O(l) Limiting reactant: Mass of UF4 = Mass of H20 = b. NaNO3(aq) + H2SO4 (aq) → Na2SO4 (aq) + HNO3(aq) Limiting reactant: Mass of Na2SO4 = Mass of HNO3 =

Answer 1

balanced equation is 00g +4 HF Ufu+q Hao 1 Identification of limiting Reactant: roles of Ug = weight of UO2(g). Tolarn 988 of Us (almale) 1 = HF = 270.03 glhole CI holes 1ooo tondes 0.004444 roles on 4.44 rotoles 120g Mules of Lo Anong vou roles. = 0.06 roles (ob Go to ruleg. Two Reactants voy has le88 Number of rol the limiting Reactant. the is thue

ng g) produces 1 Mole of UH4 (314002969 ) I role of vog (970.03 917) produces produces ? grig of Ufy - 120g of vow grs of UFG - 1.208314.02 - 070-03 = 1.395 grs of UFC in anount of UF4 = 1:395g mass of Ho: Molar rags of Hyo: 18g/role. hole of voz (270.03 97) Produced a role of ligo (2X 18 902 120 gr of vo produces ? gn of H2O = 1.20 X2X18 gn of H₂o - 290.03 = 0.159 grot Hot & amount of Hao = 0.159 g

6 balanced equation is & Na Nogt Hasoy > (ar) (aq) Identification of cititing Nag soy of 2 H Nos (ac) (aw) Reactant: roles of Na Nose weight of dadog (9) Molar mass of Japoz (gnole 2 1.204 १५.११५ glhue a prolull holes con Hill Mitoles. roles of Hysay a 1204 I a8 glide = 0.01224 roles (on 12.24 rituales, : Anong the two Reactants Hesoy has less runber of Moleq, thus Hesoy is the limiting Reactant.

atant of Naqsoy: 1.20 g of Hy soy produces ?grs of Nasoy 2 1920x142 2 1090 grs of Naz Soy z 17 38 ghs of Maly say ... Anont of Naz soy = 1.738 gn anoont of Heaza I role of H₂ soy (9892) produces erodes of Hug (2x630ne 1,20 g of H₂ soy produces ? 9hs of Hoog = 120x2x63 ahol Hoog = 1.542 SHS of Holog