What volume of oxygen, measured at STP, can be prepared by heating 2.16g HgO?

To determine the volume of oxygen gas produced when heating 2.16g of HgO, we need to use the concept of stoichiometry.

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

2. Calculate the molar mass of HgO:
HgO: (200.59 g/mol Hg + 15.999 g/mol O) = 216.59 g/mol

3. Convert the given mass of HgO to moles:
moles = mass / molar mass = 2.16g / 216.59 g/mol = 0.01 moles

4. Use the stoichiometry of the balanced chemical equation to relate moles of HgO to moles of O2:
From the balanced equation, we see that 2 moles of HgO produce 1 mole of O2.

Therefore, moles of O2 = 0.01 moles HgO * (1 mole O2 / 2 moles HgO) = 0.005 moles O2

5. Apply the ideal gas law to determine the volume of oxygen at STP:
According to the ideal gas law, 1 mole of any gas at STP occupies 22.4 liters.

volume = moles * 22.4 L/mol = 0.005 moles O2 * 22.4 L/mol = 0.112 L

Therefore, the volume of oxygen gas produced when heating 2.16g of HgO at STP is 0.112 liters.

To determine the volume of oxygen gas produced by heating 2.16g of HgO, we will need to use the concept of stoichiometry. The balanced chemical equation for the reaction between HgO (mercury(II) oxide) and heat is as follows:

2 HgO(s) -> 2 Hg(l) + O2(g)

From the equation, we can see that it takes 2 moles of HgO to produce 1 mole of O2 gas. To find the number of moles of HgO, we need to divide the given mass (2.16g) by its molar mass.

The molar mass of HgO can be calculated by adding the atomic masses of mercury (Hg) and oxygen (O). Looking up the atomic masses:
- The molar mass of Hg is approximately 200.59 g/mol.
- The molar mass of O is approximately 16 g/mol.

Thus, the molar mass of HgO is calculated as follows:
Molar mass of HgO = (200.59 g/mol) + (16.00 g/mol) = 216.59 g/mol

Now we can calculate the moles of HgO:
moles of HgO = mass of HgO / molar mass of HgO
moles of HgO = 2.16 g / 216.59 g/mol

moles of HgO ≈ 0.009955 mol (rounded to 4 decimal places)

According to the stoichiometry of the balanced equation, 2 moles of HgO react to form 1 mole of O2. Therefore, the number of moles of O2 produced can be calculated as half of the moles of HgO.

moles of O2 = moles of HgO / 2
moles of O2 ≈ 0.009955 mol / 2

moles of O2 ≈ 0.004978 mol (rounded to 4 decimal places)

Now, to determine the volume of O2 gas at STP (Standard Temperature and Pressure), we can use the Ideal Gas Law. STP conditions are defined as 1 mole of gas occupying 22.4 liters of volume.

volume of O2 = moles of O2 * 22.4 L/mol
volume of O2 ≈ 0.004978 mol * 22.4 L/mol

volume of O2 ≈ 0.11143 L (rounded to 5 decimal places)

Therefore, the volume of oxygen gas produced by heating 2.16g of HgO at STP would be approximately 0.11143 liters.

1. Write and balance the equation.

2. Convert g HgO to mols. mols = g/molar mass.
3. Using the coefficients in the balanced equation, convert mols HgO to mols O2.
4. Now convert mols O2 to volume. 1 mol O2 occupies 22.4 L at STP.