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Given: 2LiBr + I2 → 2LiI + Br2 Calculate the mass of bromine produced when 9.033 × 1023 particles of iodine (I2) react completely. Express your answer to the correct number of significant figures.

The mass of the Br2 ? is grams.

2 Answers

3 votes

Answer:

239.7 g

Step-by-step explanation:

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User Ilya Klyuchnikov
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4 votes

Answer:

239.7 g

Step-by-step explanation:

Step 1: Write the balanced equation

2 LiBr + I₂ → 2 LiI + Br₂

Step 2: Convert the molecules of iodine to moles

We have 9.033 × 10²³ particles (molecules) of iodine. In order to convert molecules to moles, we will use the Avogadro's number: there are 6.022 × 10²³ molecules of iodine in 1 mole of iodine.


9.033 * 10^(23)molecule * (1mol)/(6.022 * 10^(23)molecule) =1.500mol

Step 3: Calculate the moles of bromine produced

The molar ratio of I₂ to Br₂ is 1:1. Then, the moles of bromine produced are 1.500 moles.

Step 4: Calculate the mass of bromine

The molar mass of bromine is 159.81 g/mol. The mass corresponding to 1.500 moles is:


1.500mol * (159.81g)/(mol) = 239.7 g

User TheWhiteRabbit
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