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.

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

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