According to the following reaction, how many grams of bromine trifluoride are needed to form 29.2 grams of fluorine gas? bromine trifluoride (g) bromine (g) + fluorine (g)

Answer 1

Hello, Jaden,

We need to balance the equation. I come up with:

2BrF₃ = Br₂ + 3F₂

We want 29.2 g of flourine gas, F₂. Using the molar mass of F₂, 38 g/mole, we can calculate this to be (29.2 g/38 g/mole), or 0.768 moles of F₂.

The balanced equation tells us we need 2 moles of BF₃ to make 3 moles of F₂, a (2 BF₃/3 F₂) molar ratio.

To obtain 0.768 moles of F₂, we therefore need:

(0.768 moles F₂)*(2 BF₃/3 F₂) = 0.512 moles BrF₃

Convert this to grams by multiplying by the molar mass of BrF₃, 136.9 g/mole. This says we need 70.1 grams of BrF₃ to make 29.2 grams of F₂.

Bob