Use this equation to answer the questions below: Br2 + Cl2 → 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 318.5 g Cl2 react with excess Br2?
c. How many grams of Br2 are needed to react with 3.36 × 1025 molecules Cl2?
Br2 + Cl2 → 2BrCl. Assuming the reaction occurs above 99.9%, then
a. 2.74 mols Cl2 x (2 mols BrCl/1 mol Cl2) = 2.74 x 2 = ?
b. mols Cl2 =grams/molar mass = 318.5/71 = about 5 but you should be more accurate.
Then 5 mols Cl2 x 2 (from part a) = 10 mols BrCl.Then grams BrCl = mols BrCl x molar mass BrCl = ? Note the about phrase. I estimated. Close but not exactly accurate.
c. Convert 3.36E25 molecules to moles. 3.36E25/6.02E23 = mols Cl2. Then follow the procedure in b.
Post your work if you get stuck.
To answer these questions, we will use the balanced equation Br2 + Cl2 → 2BrCl.
a. To find the number of moles of BrCl formed, we first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.
In this case, Br2 is in excess while Cl2 is given as 2.74 mol. Since the stoichiometric ratio between Cl2 and BrCl is 1:2, we can calculate the number of moles of BrCl formed as follows:
2.74 mol Cl2 × (2 mol BrCl / 1 mol Cl2) = 5.48 mol BrCl
Therefore, 5.48 moles of BrCl will form when 2.74 mol of Cl2 react with excess Br2.
b. To find the mass of BrCl formed, we need to calculate the moles of BrCl formed using the same stoichiometric ratio.
318.5 g Cl2 × (1 mol Cl2 / 70.906 g Cl2) × (2 mol BrCl / 1 mol Cl2) × (79.904 g BrCl / 1 mol BrCl) = 672.68 g BrCl
Therefore, 672.68 grams of BrCl will form when 318.5 grams of Cl2 react with excess Br2.
c. To find the mass of Br2 needed, we need to calculate the number of moles of Cl2 and then use the stoichiometric ratio.
3.36 × 10^25 molecules Cl2 × (1 mol Cl2 / 6.0221 × 10^23 molecules Cl2) × (1 mol Br2 / 1 mol Cl2) × (159.808 g Br2 / 1 mol Br2) = 8.39 × 10^6 g Br2
Therefore, approximately 8.39 × 10^6 grams of Br2 are needed to react with 3.36 × 10^25 molecules of Cl2.
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.