Question

The reaction 6ClO2 (g) + 2BrF3 (l) --> 6ClO2F (s) + Br2 (l) is carried out with 12 mol of ClO2 and 5 mol of BrF3. a) Identify the excess reactant. b) Estimate how many moles of each product will be produced and how many moles of the excess reactant will remain. c) Which reactant is limiting?

Answer 1

6ClO2 (g) + 2BrF3 (l) --> 6ClO2F (s) + Br2 (l)

a) according to balanced reaction

6 moles ClO2 reacts with 2 moles BrF3

12 moles moles ClO2 reacts with 12 x 2 / 6 = 4.0 moles BrF3

but we have 5 moles of BrF3'so exess reagent = BrF3

C) limiting reagent = ClO2

b) moles of exess reagent left = 5 - 4 = 1

6 moles of ClO2 forms 6moles of 6ClO2F

12 moles of ClO2 gives 12 x 6 / 6 = 12

moles of ClO2F formrd = 12

6 moles of CLO2 forms 1 mole Br2

12 moles of ClO2 forms 12 x 1 / 6 = 2.0

moles of Br2 formedd= 2.0