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If 28.5 grams of bromine trifluoride react to form 16.6 grams of bromine, how many grams of fluorine must simultaneously be formed?

A) 11.9 grams
B) 28.5 grams
C) 16.6 grams
D) 40.7 grams

User Sebszyller
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1 Answer

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Final answer:

The mass of fluorine formed in the reaction is 11.9 grams, which is derived from subtracting the mass of bromine from the initial mass of bromine trifluoride.

Step-by-step explanation:

To solve this question, we should apply the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction. Therefore, the mass of the reactants must equal the mass of the products. If 28.5 grams of bromine trifluoride (BrF3) reacts and 16.6 grams of bromine (Br2) is formed, we can calculate the mass of fluorine (F2) produced.

The mass of the fluorine can be found by subtracting the mass of bromine from the initial mass of bromine trifluoride:

Mass of fluorine = Mass of bromine trifluoride - Mass of bromine

Mass of fluorine = 28.5 grams - 16.6 grams

Mass of fluorine = 11.9 grams

Therefore, the mass of fluorine formed is 11.9 grams, which corresponds to option A.

User DadaB
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