To answer these questions, we will use the given chemical equation: Br2 + Cl2 → 2BrCl.
a. How many moles of BrCl form when 2.74 mol Cl2 react with excess Br2?
From the balanced equation, we can see that the mole ratio between Cl2 and BrCl is 1:2. This means that for every 1 mol of Cl2, 2 mol of BrCl will be formed.
Given that we have 2.74 mol of Cl2, we can use the mole ratio to calculate the moles of BrCl:
2.74 mol Cl2 * 2 mol BrCl / 1 mol Cl2 = 5.48 mol BrCl
Therefore, when 2.74 mol of Cl2 reacts, 5.48 mol of BrCl will form.
b. How many grams of BrCl form when 318.5 g Cl2 react with excess Br2?
To solve this question, we need to convert grams of Cl2 to moles using its molar mass. The molar mass of Cl2 is approximately 70.9 g/mol.
Given that we have 318.5 g of Cl2, we can use the molar mass to calculate the moles of Cl2:
318.5 g Cl2 * 1 mol Cl2 / 70.9 g Cl2 = 4.49 mol Cl2
Now, we can use the mole ratio from the balanced equation to calculate the moles of BrCl. From the equation, we know that the mole ratio between Cl2 and BrCl is 1:2.
4.49 mol Cl2 * 2 mol BrCl / 1 mol Cl2 = 8.98 mol BrCl
To find the grams of BrCl, we can multiply the moles by the molar mass of BrCl. The molar mass of BrCl is approximately 187.8 g/mol.
8.98 mol BrCl * 187.8 g BrCl / 1 mol BrCl = 1,687.24 g BrCl
Therefore, when 318.5 g of Cl2 reacts, approximately 1,687.24 g of BrCl will form.
c. How many grams of Br2 are needed to react with 3.36 × 10^25 molecules of Cl2?
To solve this question, we need to convert the number of molecules of Cl2 to moles. We know that 1 mole of any substance contains 6.022 × 10^23 molecules (Avogadro's number).
Given that we have 3.36 × 10^25 molecules of Cl2, we can use the following conversion:
3.36 × 10^25 molecules Cl2 * 1 mol Cl2 / 6.022 × 10^23 molecules Cl2 = 55.86 mol Cl2
Now, we can use the mole ratio from the balanced equation to calculate the moles of Br2. From the equation, we know that the mole ratio between Cl2 and Br2 is 1:1.
Thus, 55.86 mol of Br2 are needed to react with 55.86 mol of Cl2.
To find the grams of Br2, we can multiply the moles by the molar mass of Br2. The molar mass of Br2 is approximately 159.8 g/mol.
55.86 mol Br2 * 159.8 g Br2 / 1 mol Br2 = 8,930.43 g Br2
Therefore, approximately 8,930.43 grams of Br2 are needed to react with 3.36 × 10^25 molecules of Cl2.