Question

In a bromine-producing plant, how many liters of gaseous elemental bromine at 277?C and 0.935 atm are formed by the reaction of 450. g of sodium bromide and 205 g of sodium bromate in aqueous acid solution? (Assume no Br2 dissolves ad that sulfuric acid is present in large excess.)

5NaBe+NaBrO3+3H2SO4 -> 3 Br2+3Na2SO4+3H2O

Answer 1

NaBr mass = 450 g

NaBr molar mass = 102.89 g/mol

NaBr moles = mass / molar mass

                    = 450 / 102.89

                    = 4.37

NaBrO3 mass = 205 g

NaBrO3 molar mass = 150.89 g /mol

NaBrO3 moles = 205 / 150.89

                         = 1.36

5NaBr +   NaBrO3     + 3H2SO4 -------------------> 3 Br2       +   3Na2SO4 +   3H2O

5mol             1mol

4.37mol       1.36mol

here limiting reagent is NaBr

so

5 moles NaBr -----------------------> 3 moles Br2

4.37 mole NaBr -------------------> x moles Br2

x = 4.37 x 3 / 5

= 2.622

moles of Br2 (n) produced = 2.622

Temperature (T) = 277 + 273 = 550 K

pressure (P) = 0.935 atm

universal gas constant (R) = 0.0821 L-atm / mol K

PV = n RT

V = nRT/P

V = 2.622 x 0.0821 x 550 / 0.935

V = 126.63 L

volume of Br2 produced = 126.63 L