How many grams of oxygen can be produced by heating 10.0g of mercury (II) oxide?

Sulfuric acid is added to 8.10g of magnesium, how many grams of hydrogen are set free?

Answered abovd.

1. 0.74g

2. .33mol/.33g of hydrogen

To calculate the grams of oxygen produced by heating mercury (II) oxide, we need to determine the molar ratio between mercury (II) oxide and oxygen.

The balanced chemical equation for the reaction is:

2 HgO -> 2 Hg + O2

The molar mass of mercury (II) oxide (HgO) is 216.59 g/mol. Since the molar ratio between HgO and O2 is 1:1, we can calculate the grams of oxygen produced using the following formula:

Grams of oxygen = (grams of HgO) * (1 mol HgO / molar mass HgO) * (molar mass O2 / 1 mol HgO)

Let's substitute the given values:

Grams of HgO = 10.0 g
Molar mass HgO = 216.59 g/mol
Molar mass O2 = 32.00 g/mol

Grams of oxygen = (10.0 g) * (1 mol / 216.59 g/mol) * (32.00 g/mol / 1 mol)
= 14.76 g

Therefore, 10.0 grams of mercury (II) oxide can produce 14.76 grams of oxygen.

Now, let's move on to the second question.

To calculate the grams of hydrogen set free when sulfuric acid reacts with magnesium, we need to determine the molar ratio between magnesium and hydrogen.

The balanced chemical equation for the reaction is:

Mg + H2SO4 -> MgSO4 + H2

The molar mass of magnesium (Mg) is 24.31 g/mol. Since the molar ratio between Mg and H2 is 1:1, we can calculate the grams of hydrogen produced using the following formula:

Grams of hydrogen = (grams of Mg) * (1 mol Mg / molar mass Mg) * (molar mass H2 / 1 mol Mg)

Let's substitute the given values:

Grams of Mg = 8.10 g
Molar mass Mg = 24.31 g/mol
Molar mass H2 = 2.02 g/mol

Grams of hydrogen = (8.10 g) * (1 mol / 24.31 g/mol) * (2.02 g/mol / 1 mol)
= 0.665 g

Therefore, when 8.10 grams of magnesium react with sulfuric acid, approximately 0.665 grams of hydrogen are set free.

To calculate the number of grams of oxygen that can be produced by heating 10.0g of mercury (II) oxide, we need to use the molar mass and balanced chemical equation.

1. Look up the molar mass of mercury (II) oxide (HgO):
- Molar mass of Hg = 200.59 g/mol
- Molar mass of O = 16.00 g/mol
- Total molar mass of HgO = 200.59 + 16.00 = 216.59 g/mol

2. Write and balance the chemical equation for the reaction:
- 2 HgO(s) -> 2 Hg(l) + O₂(g)

3. Determine the number of moles of HgO:
- Moles of HgO = mass of HgO / molar mass of HgO
- Moles of HgO = 10.0 g / 216.59 g/mol

4. According to the balanced equation, each mole of HgO produces 1 mole of O₂:
- Moles of O₂ produced = moles of HgO

5. Calculate the mass of O₂ produced:
- Mass of O₂ = moles of O₂ produced * molar mass of O
- Mass of O₂ = moles of HgO * molar mass of O

For the second question:

To calculate the number of grams of hydrogen set free when sulfuric acid is added to 8.10g of magnesium, we'll follow a similar process:

1. Look up the molar mass of magnesium (Mg):
- Molar mass of Mg = 24.31 g/mol

2. Write and balance the chemical equation for the reaction:
- Mg(s) + H₂SO₄(aq) -> MgSO₄(aq) + H₂(g)

3. Determine the number of moles of magnesium (Mg):
- Moles of Mg = mass of Mg / molar mass of Mg
- Moles of Mg = 8.10 g / 24.31 g/mol

4. According to the balanced equation, each mole of Mg produces 1 mole of H₂:
- Moles of H₂ produced = moles of Mg

5. Calculate the mass of hydrogen (H₂) produced:
- Mass of H₂ = moles of H₂ produced * molar mass of H₂
- Mass of H₂ = moles of Mg * (2 g/mol)