Answer:
a) Br2 + 3F2 → 2BrF3
<u>b) Br2 is the limiting reactant</u>.
c) There will be formed 86.3 grams of BrF3
d) There will remain <u>0.326 moles of F2 = 12.97 grams</u>
<u />
Explanation:
Step 1: The balanced equation
Br2 + 3F2 → 2BrF3
Step 2: Given data
Mass of Bromine = 63grams
Mass of fluorine = 60 grams
percent yield = 80%
Molar mass of bromine = 79.9 g/mol =
Molar mass of fluorine = 19 g/mol
Molar mass of bromine trifluoride = 136.9 g/mol
Step 3: Calculating moles
Moles Br2 = 63 grams / (2*79.9)
Moles Br2 = 0.394 moles
Moles F2 = 60 grams / (2*19.9)
Moles F2 = 1.508
For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3
<u>Br2 is the limiting reactant</u>. It will completely be consumed (0394 moles).
There will react 3*0.394 = 1.182 moles of F2
There will remain 1.508 - 1.182 = <u>0.326 moles of F2 = 12.97 grams</u>
Step 5: Calculate moles of BrF3
For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3
So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced
Step 6: Calculate mass of BrF3
mass = Moles * Molar mass
mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield
Step 7: Calculate actual yield
% yield = 0.80 = actual yield / theoretical yield
actual yield = 0.80 * 107.88 grams = 86.3 grams
actual yield = <u>86.3 grams</u>
<u />
<u />
