Question

15 Determine the theoretical yield of H2S lin moles) if 16 mol Ales, and 16 mol H₂O are reacted according to the following balanced reaction. A possibly useful molor mass is AS, E150,17 g/mol Al, S, (s) + 6H2O10-2 Alco),(s)+3H,SA) ) 48. HAS (B) 16 mol HS c)32mol H,S 8.0 mol H.S (E) 24 md H₂S 16 A 14.01g Sample of Ng reacts with 3.024 of Ho, to form ammonia (AH2). If ammonia is the only product, What mass of ammonia is formed? (determine limited reactant first) 6) 23.079 (B) 1.109 (C)3.029 () 17.03g E) 14.019 I dentify acetic acid. [(A) nonelectrolyte B)weak electrolyte, weak acid © weak electrolyte, Strong acid D) Strong electrolyte, Strong acid E) Strong electrolule, weak oud 1 (8) Identify HCI. (A) weak electrolyte, strong acid | (B) nonelectrolute C weakelectrolyte, weak acid D) Strong electrolyte, strong acid | (E) Strong electrolyte, weak acid

Answer 1

AL 5. Given 16 mol Al₂S₃ and 16 mol to Abysz (s) + 6406) → 2 Al(OH)3 (s) + 3 Hy$ (9) Mole Tatto Tatto! . : 3 For 16 mol of Absz GX16 = 96 mol of tho is required but lb molen Hyo Ps only given. So tho is limiting Reagent and Alfzis excess reagent & For Imole of Abaz , 6 mol of H₂O is reauired (116) mole of Alafz, 16 mol of tho is required 1. 22.66 moles * Also 6 mol of H20 gives a moles of ths 16 ml of tho will give dx = Emol of Ihs option D is correct 6. Given 14:0g of My and 3:02 g of Hy - N₂ + 3H₂ → 2NH₂ moles 13 Mass t mass En er 24 10489 RX 17.03 $34.06 28:013 g of My require a g of H2 1.4914001 9 of My Teamuite 3.003 1 = 14.01X6.048 -3.079) 28.0134 1 given masse So the amount 8 NHz formed is jer bi s 1101*34.06 17.0391 i lopbron pts correct 28.0134 7 AceHtc acid is Weak electrolyte and weak and because it dissociation Constant is small. So there will be few ions in Solution to conduct. So 9tis Weak electrolyte and weak acid) o Hd Ps throng electrolyte and strong acid because its dissociation constant is high, so there will be more pons in Selution to conduct.