Question

Consider the reaction: Br2(g) + Cl2(g) → 2BrCl(g) Given an initial mass of 14.21 g Br2, an excess of Cl2, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of BrCl produced by the reaction.

Answer 1

Answer:

20.52 grams of BrCl will be produced

Explanation:

Step 1: Data given

Mass of Br2 = 14.21 grams

Cl2 is in excess

Molar mass of Br2 = 159.81 g/mol

Molar mass BrCl = 115.36 g/mol

Step 2: The balanced equation

Br2(g) + Cl2(g) → 2BrCl(g)

Step 3: Calculate moles Br2

Moles Br2 = mass Br2 / molar Br2

Moles Br2 = 14.21 grams / 159.81 g/mol

Moles Br2 =0.08892 moles

Step 4: Calculate moles BrCl

For 1 mol Br2 we need 1 mol Cl2 to produce 2 moles BrCl

For 0.08892 moles Br2we'll have 2*0.08892 = 0.17784 moles BrCl

Step 5: Calculate mass BrCl

Mass BrCl = moles BrCl * Molar mass BrCl

Mass BrCl = 0.17784 moles * 115.36g/mol

Mass BrCl = 20.52 grams

20.52 grams of BrCl will be produced

Answer 2

Answer:

20.5 g of bromine chloride are produced in the reaction

Explanation:

In the reaction we see that 1 mol of bromine react to 1 mol of chlorine to produce 2 moles of bromine chloride

Equation: Br₂(g) + Cl₂(g) → 2BrCl(g)

As we assumed the chlorine as the excess, we convert the mass of bromines to moles → 14.21 g . 1mol / 159.8 g = 0.0889 moles

In the stoichiometry 1 mol of bromine produces 2 moles of chloride.

Therefore, 0.0889 moles of Br₂ will produce (0.0889 . 2) /1 = 0.178 moles of chloride. Let's convert the moles to mass, to get the answer:

0.178 moles BrCl . 115.35g /1mol = 20.5 g