Question

Metallic sodium reacts vigorously with liquid bromine in the following reaction: 2Na + Br₂ -----> 2NaBr

1 kg of Na is brought into contact with 3 kg of liquid bromine. Presuming that the reaction is quantitative and proceeds to completion, determine the limiting reagent and the quantity of NaBr formed. Determine the amount of excess reagent remaining after the reaction is complete.

Answer 1

molar mass of Na = 22.989 gm/mole then 2 mole of Na = 45.978 gm

molar mass of Br₂ = 159.808 gm/mole

1kg = 1000 gm

3kg = 3000 gm

According to reaction 2 mole of Na react with 1 mole of Br₂ that mean 45.978 gm of Na react with 159.808 gm of Br then to react with 1000 gm of Na required Br = 1000 159.808 / 45.978 = 3475.74 gm of Br

but Br given only 3000 gm therefore Br₂ is limiting reagent.

molar mass of NaBr = 102.894 gm / molel then 2 mole of NaBr = 205.788 gm

according to reaction 1 mole of Br₂ produce 2 mole of NaBr that mean 159.808 gm of Br₂ produce 205.788 gm of NaBr then 3000 gm of Br₂ produce = 3000 205.788 / 159.808 = 3863.16 gm of NaBr

NaBr produced = 3863.16 gm = 3.86316 kg

According to reaction 2 mole of Na react with 1 mole of Br₂ that mean 45.978 gm of Na react with 159.808 gm of Br then to react with 3000 gm of Br₂ required Na = 3000 45.978 / 159.808 = 863.12 gm of Na

but Na given = 1000 gm

Excess Na remain after reaction complete = 1000 - 863.12 = 136.88 gm