Question

I found the Delta H rxn to be -1376 KJ . I'm confused on the steps to find my limiting reactant and grams of BCl3.

Question How many grams of BClz will be produced if 55.36 g of B, Ho reacted with an unknown mass of Cl, in which the reaction released 688.0 KJ of heat? Which is the limiting reactant? BH6 + 6Cl2 → 2BCl2 + 6HCI AHrxn = ? Given: 1/2H2 + 1/2012 B H6 + 6H2O BCl3 + 3H2O → HCI → 2H3BO2 + 6H2 → H3BO3 + 3HCI AH = -92.3 KJ AH = - 493.4 KJ AH = - 112.5 KJ 6-58

Answer 1

BzHG + GCI 2BCl3 + 6HCl Ohran Molar Mass of 132H6 » 2(10.811) + 6(1) = 27.622 ohon = 688 kJ => so to release oHorn = 688 k] heat required amount of B2M6 = 27.622 x 688 1376 WB2Ho = 13.811 and liven amount of B2 H is = 55-3ogm - means B2 HG is in excess so limiting reagent = C12 Ang. AO Produced BCl3 nocl3 = 2 x oH given) OHO = 2 x688 1376 MBCl3 = Imol MM = 10.811 +3 (35.45) =) 117.161 gm imal So W Buiz - A x 117-161) CUBCI3 = 117.16 gm Ang

Target equation Bato +6012 2BCl3 + 6HCl; Shan? Cuiven oyn Ž H2 + { cl2 — HCL; OH, = -92-3 k3 B2 HG + G H2O → 2H₂ BO₂ + 6H₂; OH₂ =- OH, = -493.4K] BCl3 + 3H2O → H3 B O2 + 3HCI ; OH, = -112-SKJ do following action to get the ternget ry eqh ② + 12 eqho - axeqh ® B2 H6 + 6H₂O 2H₂ BO + 6H, OH₂ = -493.4K] 6H2 +6Cl2 - 12HCI ; OH = -110706 k] 2 H3B0% + 6HCI — 2BC12 +6H20; OH' = 225 kJ B2 HG + 6Cl2 → 2Bl3 +GHC ; alian Olegan = OH2 + oH; + Oms! : - 493.3 - 1 107,6 + 225 ohman = -1375.9 kJ/mol