1.) Given: 2H2O2 = 2H2O + O2a.

a.) What mass of water is produced when 5.0 mol O2 is produced?

From the balanced chemical equation, we can see that 1 mole of O2 is produced for every 2 moles of H2O. Therefore, if 5.0 mol of O2 is produced, then 10.0 mol of H2O must also be produced.

The molar mass of H2O is approximately 18.015 g/mol, so:

mass of water = 10.0 mol × 18.015 g/mol = 180.15 g of water

Therefore, 180.15 grams of water are produced when 5.0 moles of O2 are produced.

2.) Given: 3Fe + 4H2O = 4H2 + Fe3O4a.

a.) If 625 g of Fe3O4 is produced on the reaction, how many moles of hydrogen are produced at the same time?

b.) How many moles of iron would be needed to generate 27 g of hydrogen?

a.) To determine the number of moles of hydrogen produced, we need to use the stoichiometry of the balanced chemical equation. From the equation, we can see that 3 moles of Fe produce 4 moles of H2. Therefore, we can set up a proportion:

3 mol Fe / 4 mol H2 = x mol Fe3O4 / 625 g Fe3O4

x = 3/4 * (625 g Fe3O4 / molar mass of Fe3O4) = 3/4 * (625 g / 231.535 g/mol) ≈ 6.45 mol

Therefore, 6.45 moles of hydrogen are produced when 625 g of Fe3O4 is produced.

b.) To determine how many moles of iron are needed to produce 27 g of hydrogen, we can use the same stoichiometry as before. From the balanced equation, we know that 3 moles of Fe produce 4 moles of H2. So, we can set up a proportion:

3 mol Fe / 4 mol H2 = x mol Fe / 27 g H2

x = 3/4 * (27 g H2 / molar mass of H2) = 3/4 * (27 g / 2.016 g/mol) ≈ 10.1 mol

Therefore, approximately 10.1 moles of iron would be needed to generate 27 g of hydrogen.

3.) Given: MnO2 + 4HCl → MnCl2 + 2H2O + Cl2a.

a.) Calculate the mass of MnO2 needed to produce 25 g of Cl2b.

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

a.) From the balanced chemical equation, we can see that 1 mole of MnO2 reacts with 4 moles of HCl to produce 1 mole of Cl2. Therefore, we can set up a proportion:

1 mol MnO2 / 4 mol HCl = x mol MnO2 / 1 mol Cl2

x = (1/4) * (1 mol Cl2 / 1 mol MnO2) = 0.25 mol MnO2

The molar mass of MnO2 is approximately 86.94 g/mol, so:

mass of MnO2 = 0.25 mol × 86.94 g/mol = 21.74 g MnO2

Therefore, 21.74

b.) From the balanced chemical equation, we can see that 1 mole of MnO2 reacts with 4 moles of HCl to produce 1 mole of Cl2 and 1 mole of MnCl2. Therefore, we can set up a proportion:

1 mol MnO2 / 4 mol HCl = 1 mol MnCl2 / 1 mol Cl2

So, 1 mole of MnCl2 is produced for every mole of Cl2.

We are given that 0.091 g of Cl2 is generated. The molar mass of Cl2 is approximately 70.906 g/mol, so:

moles of Cl2 = 0.091 g / 70.906 g/mol ≈ 0.001284 mol Cl2

Since 1 mole of MnCl2 is produced for every mole of Cl2, the number of moles of MnCl2 produced is also 0.001284 mol. The molar mass of MnCl2 is approximately 125.844 g/mol, so:

mass of MnCl2 = 0.001284 mol × 125.844 g/mol ≈ 0.162 g MnCl2

Therefore, approximately 0.162 g of MnCl2 is produced when 0.091 g of Cl2 is generated.

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