Question

If 22.8 grams of bromine and 10.1 grams of chlorine combine to form bromine monochloride, how many grams of bromine monochloride must form?

Answer 1

The balanced chemical reaction between bromine and chlorine,

Br₂(g) + Cl₂(g) 2BrCl(g)

Given,

Mass of Br₂ = 22.8 g

Mass of Cl₂ = 10.1 g

Calculating the number of moles of Br₂ and Cl₂ from the given masses,

= 22.8 g Br₂ x ( 1 mol / 159.808 g)

= 0.1427 mol Br₂

Similarly,

= 10.1 g Cl₂ x ( 1 mol / 70.9060 g)

= 0.1424 mol Cl₂

Now, calculating the number of moles of Cl₂ required to react completely with given moles of Br_2^.

= 0.1427 mol Br₂ x ( 1 mol Cl₂/ 1 mol Br₂)

= 0.1427 mol Cl₂ required to react completely with the given moles of Br₂(0.1427 mol)

Similarly,

calculating the number of moles of Br₂ required to react completely with given moles of Cl_2^.

= 0.1424 mol Cl₂ x ( 1 mol Br₂/ 1 mol Cl₂)

= 0.1424 mol Br₂ required to react completely with the given moles of Cl₂(0.1424 mol)

since moles of Cl₂ required to react completely with the given moles of Br₂ are more, thus Cl₂ is the limiting reactant.

Now, using the moles of Cl₂ and calculating the number of moles of BrCl formed,

= 0.1424 mol Cl₂ x ( 2 moles of BrCl / 1 mol Cl₂)

= 0.2849 mol BrCl formed

Converting moles to grams,

= 0.2849 mol BrCl x (115.357 g / 1 mol)

= 32.9 g BrCl must form, [3 S.F]