Question

Bromine is commercially prepared from natural brines coming from wells in Michigan and Arkansas by the reaction 2 NaBr(aq) + Cl2(g) ® 2 NaCl(aq) + Br2(l) How many grams of bromine can be prepared by reaction of 5.581 g of NaBr with 2.284 g of Chlorine?

Answer 1

First calculate limiting reactant

molar mass of NaBr = 102.894 gm/mole then 5.581 gm NaBr = 5.581 / 102.894 = 0.05424 mole

molar mass of Cl₂ = 70.9 gm/mole then 2.284 gm of Cl₂ = 2.284 / 70.9 = 0.03221 mole

According to balanced chemical reaction 2 mole of NaBr react with 1 mole of Cl₂ therefore to react with 0.03221 mole of Cl₂ required NaBr = 0.03221 X 2 = 0.06443 mole of NaBr but NaBr given only 0.05424 mole therefore NaBr is limiting reactant

According to reaction 2 mole of NaBr produce 1 mole of Br₂ then 0.05424 mole of NaBr produce 0.05424 mole of Br₂

molar mass Br₂ = 159.808 gm/mol then 0.05424 mole of Br₂ = 159.808 X 0.05424 = 8.67 gm

8.67 gm Br₂ can be prepared by reaction of 5.581 g of NaBr with 2.284 g of Chlorine.