Question

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.

Answer 1

Answer:

239.7 g

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.

$9.033 \times 10^{23}molecule \times \frac{1mol}{6.022 \times 10^{23}molecule} =1.500mol$

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:

$1.500mol \times \frac{159.81g}{mol} = 239.7 g$