According to the following reaction, how many grams of bromine trifluoride are needed to form 21.5 grams of fluorine gas?

Question

According to the following reaction, how many grams of bromine trifluoride are needed to form 21.5 grams of fluorine gas?

bromine trifluoride (g) bromine (g) + fluorine (g)

Answer 1

According to the following reaction, how many grams of bromine trifluoride are needed to form 21.5 g fluorine gas? bromine trifluoride (g)  bromine (g) + fluorine (g)

You must first write the formulas for the compounds (called the skeleton equation), then balance the equation.

Skeleton equation is BrF₃ Br₂ + F₂ (bromine and fluorine are diatomic elements)

Balanced equation is 2 BrF₃ Br₂ + 3 F₂

Asked / given ................? g ...................21.5 g

Molar masses..........136.9 ^g/_mol ............38.0 ^g/_mol

Begin with the quantity you are given in the problem, 21.5 g F₂

Overall, the steps are:

1 convert mass F₂ to mol F₂ using the molar mass of F₂;

2 convert mol F₂ to mol BrF₃ using the coefficients from the balanced equation (stoichiometry)

3 convert mol BrF₃ to g BrF₃ using the molar mass of BrF₃

21.5 g F₂ x ^{1 mol F2}/_{38.0 g F2} = answer #1 (unit is mol F₂)

answer #1 (mol F₂ ) x ^{2 mol BrF3}/_{3 mol F2} = answer #2 (unit is mol BrF₃)

answer #2 (mol BrF₃ ) x 136.9 g BrF₃ /1 mol BrF₃ = answer #3 (unit is g BrF₃)