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barxatty [35]
3 years ago
14

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the

limiting reactant? c) Determine how many grams of the product will form if the percent yield of the reaction is 80%. d) Determine how many grams of the excess reactant is left unreacted.
Chemistry
1 answer:
fiasKO [112]3 years ago
7 0

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>

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