Chlorine gas can be produced in the laboratory by adding concentrated hydrochloric acid to manganese(IV) oxide in the following reaction..

MnO2(s)+ 4HCl(aq)-->MnCl2(aq)+ 2H2O(l)+ Cl2(g)

A. Calculate the mass of MnO2 needed to produce 25.0g of Cl2

B. What mass of MnCl2 is produced when 0.091 g of Cl2 is generated?

0.16g MnCl2

Le n.0 du manganese est passe

A. To calculate the mass of MnO2 needed to produce 25.0g of Cl2, we can use the balanced equation to determine the molar ratio between MnO2 and Cl2.

From the balanced equation, we can see that 1 mole of MnO2 produces 1 mole of Cl2. We need to calculate the moles of Cl2 produced and then convert it to the mass of MnO2.

1 mole of Cl2 = 35.453g (molar mass of Cl2)
25.0g of Cl2 = (25.0g Cl2) × (1 mol Cl2 / 35.453g Cl2) = 0.705 moles of Cl2

Since 1 mole of MnO2 produces 1 mole of Cl2, the same number of moles of MnO2 is needed to produce 0.705 moles of Cl2.

Therefore, the mass of MnO2 needed can be calculated as:
Mass of MnO2 = (0.705 moles Cl2) × (86.936g MnO2 / 1 mol Cl2)
Mass of MnO2 = 61.5g

So, 61.5 grams of MnO2 is needed to produce 25.0 grams of Cl2.

B. To determine the mass of MnCl2 produced when 0.091g of Cl2 is generated, we can use the balanced equation and the molar ratio between Cl2 and MnCl2.

From the balanced equation, we can see that 1 mole of Cl2 produces 1 mole of MnCl2. We need to calculate the moles of Cl2 produced and then convert it to the mass of MnCl2.

1 mole of Cl2 = 35.453g (molar mass of Cl2)
0.091g of Cl2 = (0.091g Cl2) × (1 mol Cl2 / 35.453g Cl2) = 0.00257 moles of Cl2

Since 1 mole of Cl2 produces 1 mole of MnCl2, the same number of moles of MnCl2 is produced as the moles of Cl2.

Therefore, the mass of MnCl2 produced can be calculated as:
Mass of MnCl2 = (0.00257 moles Cl2) × (125.844g MnCl2 / 1 mol Cl2)
Mass of MnCl2 = 0.323g

So, 0.323 grams of MnCl2 is produced when 0.091 grams of Cl2 is generated.

To solve these problems, you'll need to use stoichiometry, which relates the amounts (in moles) of reactants and products in a balanced chemical equation. Here's how to approach each part of the problem:

A. Calculate the mass of MnO2 needed to produce 25.0g of Cl2:

Step 1: Convert the given mass of Cl2 to moles.
Use the molar mass of Cl2, which is approximately 35.45 g/mol, to convert grams to moles:
moles of Cl2 = (mass of Cl2) / (molar mass of Cl2)

Step 2: Use the stoichiometry from the balanced equation to relate moles of Cl2 to moles of MnO2.
From the balanced equation, it is clear that the ratio of Cl2 to MnO2 is 1:1. Therefore, the moles of MnO2 needed are equal to the moles of Cl2.

Step 3: Convert moles of MnO2 to mass.
Use the molar mass of MnO2, which is approximately 86.94 g/mol, to convert moles to grams:
mass of MnO2 = (moles of MnO2) * (molar mass of MnO2)

B. What mass of MnCl2 is produced when 0.091 g of Cl2 is generated?

Step 1: Convert the given mass of Cl2 to moles.
Use the molar mass of Cl2 to convert grams to moles.

Step 2: Use the stoichiometry from the balanced equation to relate moles of Cl2 to moles of MnCl2.
From the balanced equation, it is clear that the ratio of Cl2 to MnCl2 is 1:1. Therefore, the moles of MnCl2 produced will be equal to the moles of Cl2 generated.

Step 3: Convert moles of MnCl2 to mass.
Use the molar mass of MnCl2 to convert moles to grams.

By following these steps, you should be able to calculate the masses of MnO2 and MnCl2 in each case.