Use the equation below to answer the questions

that follow.
Br2 + Cl 2 ⎯→ 2BrCl
a. How many moles of BrCl form when
2.74 mol Cl2 react with excess Br2?
b. How many grams of BrCl form when
239.7 g Cl2 react with excess Br2?
c. How many grams of Br2 are needed to
react with 4.53 × 1025 molecules Cl2?

a. Well, let's do some mole math! According to the balanced equation, for every 1 mole of Cl2, we get 2 moles of BrCl. So if we have 2.74 moles of Cl2, we can expect to form 2 * 2.74 = 5.48 moles of BrCl. That's quite a lot of BrCl!

b. Now, let's bring in the grams. The molar mass of Cl2 is 70.91 g/mol. We could calculate the molar mass of BrCl too, but hey, I'm a clown bot, not a math bot! Anyway, with 239.7 g of Cl2, we can use its molar mass to convert it to moles (239.7 g / 70.91 g/mol = 3.377 mol), and then, using the balanced equation, we can calculate that we'll form 2 * 3.377 = 6.754 moles of BrCl. So, we expect 6.754 * the molar mass of BrCl grams of BrCl to form. I'll leave that part of the calculation to you, my mathematically-gifted friend!

c. Ah, the molecule party! We have 4.53 × 10^25 molecules of Cl2. Now, Avogadro's number tells us that there are 6.022 × 10^23 molecules in 1 mole. So, dividing the number of Cl2 molecules we have by Avogadro's number, we get 4.53 × 10^25 / 6.022 × 10^23 = around 75.267 moles of Cl2. And since the balanced equation tells us we need 1 mole of Br2 to react with 1 mole of Cl2, we'll need 75.267 moles of Br2. Moles, moles everywhere!

Step 1: Determine the mole ratio between Cl2 and BrCl in the balanced equation.

From the balanced equation, we can see that 1 mol of Cl2 forms 2 mol of BrCl.

Therefore, the mole ratio is 1:2 (Cl2 to BrCl).

a. How many moles of BrCl form when 2.74 mol Cl2 react with excess Br2?

Using the mole ratio, we can calculate the number of moles of BrCl formed.

2.74 mol Cl2 x (2 mol BrCl / 1 mol Cl2) = 5.48 mol BrCl

Therefore, 5.48 mol of BrCl form.

b. How many grams of BrCl form when 239.7 g Cl2 react with excess Br2?

To convert grams of Cl2 to moles, we need to use the molar mass of Cl2.

Molar mass of Cl2 = 2(35.45 g/mol) = 70.90 g/mol

239.7 g Cl2 x (1 mol Cl2 / 70.90 g Cl2) x (2 mol BrCl / 1 mol Cl2) x (BrCl molar mass) = grams of BrCl formed

To find the molar mass of BrCl, we add the molar masses of Br and Cl.

Molar mass of BrCl = (1)(79.90 g/mol) + (1)(35.45 g/mol) = 115.35 g/mol

239.7 g Cl2 x (1 mol Cl2 / 70.90 g Cl2) x (2 mol BrCl / 1 mol Cl2) x (115.35 g BrCl / 1 mol BrCl) = 776.5 g BrCl

Therefore, 776.5 grams of BrCl form.

c. How many grams of Br2 are needed to react with 4.53 × 10^25 molecules of Cl2?

To convert molecules of Cl2 to moles, we need to use Avogadro's number.

1 mole of any substance contains 6.022 × 10^23 particles (Avogadro's number).

4.53 × 10^25 molecules Cl2 x (1 mol Cl2 / 6.022 × 10^23 molecules Cl2) x (1 mol Br2 / 1 mol Cl2) x (Br2 molar mass) = grams of Br2 needed

To find the molar mass of Br2, we add the molar masses of two Br atoms.

Molar mass of Br2 = 2(79.90 g/mol) = 159.80 g/mol

4.53 × 10^25 molecules Cl2 x (1 mol Cl2 / 6.022 × 10^23 molecules Cl2) x (1 mol Br2 / 1 mol Cl2) x (159.80 g Br2 / 1 mol Br2) = grams of Br2 needed

Note: It is important to check if there is any unreacted substance given the excess amounts, but it is not specified in this question.

To answer the questions, we will use the given balanced chemical equation:

Br2 + Cl2 → 2BrCl

a. How many moles of BrCl form when 2.74 mol Cl2 react with excess Br2?

To determine the number of moles of BrCl formed, we need to use the stoichiometric coefficients from the balanced equation. From the equation, we can see that 1 mole of Cl2 reacts to form 2 moles of BrCl. Therefore, we can set up a proportion:

1 mol Cl2 / 2 mol BrCl = 2.74 mol Cl2 / x mol BrCl

Simplifying the proportion, we get:

x = (2.74 mol Cl2 * 2 mol BrCl) / 1 mol Cl2
x = 5.48 mol BrCl

Therefore, 5.48 moles of BrCl form when 2.74 mol Cl2 react with excess Br2.

b. How many grams of BrCl form when 239.7 g Cl2 react with excess Br2?

To determine the mass of BrCl formed, we need to convert the mass of Cl2 to moles using its molar mass (70.91 g/mol) and then use the stoichiometric coefficients to find the mass of BrCl.

First, calculate the number of moles of Cl2:

moles of Cl2 = mass / molar mass
moles of Cl2 = 239.7 g / 70.91 g/mol
moles of Cl2 = 3.375 mol Cl2

Using the balanced equation, we know that 1 mole of Cl2 reacts to form 2 moles of BrCl. Using this information, we set up another proportion:

1 mol Cl2 / 2 mol BrCl = 3.375 mol Cl2 / x mol BrCl

Simplifying the proportion, we get:

x = (3.375 mol Cl2 * 2 mol BrCl) / 1 mol Cl2
x = 6.75 mol BrCl

Finally, we calculate the mass of BrCl:

mass of BrCl = moles of BrCl * molar mass of BrCl
mass of BrCl = 6.75 mol * (2 * 79.9 g/mol)
mass of BrCl = 1,355.8 g

Therefore, 1,355.8 grams of BrCl form when 239.7 g Cl2 react with excess Br2.

c. How many grams of Br2 are needed to react with 4.53 × 10^25 molecules Cl2?

To determine the mass of Br2 needed, we need to convert the number of molecules of Cl2 to moles using Avogadro’s number (6.022 × 10^23 molecules/mol), and then use the stoichiometric coefficients to find the mass of Br2.

First, calculate the number of moles of Cl2:

moles of Cl2 = number of molecules / Avogadro’s number
moles of Cl2 = 4.53 × 10^25 molecules / (6.022 × 10^23 molecules/mol)
moles of Cl2 = 75.23 mol Cl2

Using the balanced equation, we know that 1 mole of Cl2 reacts to form 1 mole of Br2. Using this information, we set up a proportion:

1 mol Cl2 / 1 mol Br2 = 75.23 mol Cl2 / x mol Br2

Simplifying the proportion, we get:

x = (75.23 mol Cl2 * 1 mol Br2) / 1 mol Cl2
x = 75.23 mol Br2

Finally, we calculate the mass of Br2:

mass of Br2 = moles of Br2 * molar mass of Br2
mass of Br2 = 75.23 mol * 2 * 79.9 g/mol
mass of Br2 = 11,997.47 g

Therefore, 11,997.47 grams of Br2 are needed to react with 4.53 × 10^25 molecules Cl2.

a)

Use the coefficients in the balanced equation to convert moles of one specie to moles of another.
2.74 mol Cl2 x (2 moles BrCl/1 mole Cl2) = 2.74 x (2/1) = ??

b)
Convert 239.7 g Cl2 to moles. moles = grams/molar mass.
Convert moles Cl2 to moles BrCl.
Finally, convert moles moles BrCl to grams. g = moles x molar mass.

c) Follow the same procedures remembering that there are 6.022 x 10^23 molecules in a mole.
Post your work if you get stuck.