Question

1.- Calculate the % yield when you react 20mg of KBrO3 and produce 5 mg of KBr, assume all other reagents are in excess. Balance the equation, Indicate the reactants and the products. (KBrO: Molecular mass: 166.999 g/mol, KBr Molecular mass: 119.002 g/mol) KBrO; + KI + HBr → KBr + H2O a) If you start the reaction with 20mg of KBro; and 20mg of Kl (molecular mass=166.0028 g/mol) with an excess of HBr what is the limiting reagent? b) How many atoms are in the 5 mg of KBr? c) If the reaction produce AH=-2456kj. Is the reaction endothermic and exothermic? d) How many atoms of Oxygen are in 20mg of KBrO3?

Answer 1

KBroz +6KI +GHBY >7KBr +34,0 + 3 12 a) moles of kBroz = 20x16* . 166.999 almok → tul98x10 moes moles of KI = 20x103 g 3 1.204x10h noles 166.0028 gimol :) I mole of kBroz giver a t moles of kBr 1.1988154 moles of KBrog gives I 8.386x10' moles of KBY ii) 6 moles of kI gives y moles of Ker 1:204x16' moles of ke gives TX11204 x16 11405x104 moles of KBY KI is the limiting reagent. b) moles of KBr = 5x1089 , 4r2 x105 ndes 119.002 g/mol 23 KBr contain 2x6.023x10 atoms 10.2x105 moles of KBy contains + 2x6.023x10x4,2x10.5 20 2.73 x16 alors 2OXIA C) AH = -24 56 KJ, so the reaction is exothermie in nalüere. d) moles of kBros - 120x162 , +.198 xió* roles I mole of kBrog giver contains) → 3x6.023x16" alimas 1.198x104mdes of kBrog contains -> 3x61023x602 1.198x104 166.999 gimo > 2.16 x 102 atoms of oxygen