How do I solve this problem?

According to the following reaction, how many grams of mercury(II) oxide are
needed to form 22.4 grams of oxygen gas?

mercury(II) oxide (s) mercury (l) + oxygen (g)grams mercury(II) oxide

Balance the equation first.

Then, change the 22.4 g O2 to moles.

Look at the coefficents in the balanced equation. That tells you how much mass of mercury(II)oxide is needed. I will be happy to critique your work, if needed.

Thanks... I don't understand, but I'm actually trying to help my daughter...hope this helps her... I have a couple more questions that I'm going to post.

To solve this problem, you need to use stoichiometry. Stoichiometry is a method used to determine the quantitative relationships between reactants and products in a chemical reaction.

To solve the problem, follow these steps:

Step 1: Write and balance the chemical equation.

The given chemical equation is:
mercury(II) oxide (s) -> mercury (l) + oxygen (g)

Step 2: Determine the molar mass of the substances involved.

The molar mass of mercury(II) oxide (HgO) is calculated by adding the molar masses of mercury (Hg) and oxygen (O). The molar mass of Hg is 200.59 g/mol, and the molar mass of O is 16.00 g/mol. So, the molar mass of HgO is 200.59 + 16.00 = 216.59 g/mol.

Step 3: Use stoichiometry to find the amount of HgO required.

According to the balanced equation, 1 mole of mercury(II) oxide reacts to form 1 mole of oxygen gas. This means that the molar ratio between HgO and O2 is 1:1.

To find the amount of HgO required to form 22.4 grams of oxygen gas, you need to convert the grams of O2 to moles using the molar mass of O2 (32.00 g/mol).

mols of O2 = given mass (22.4 g) / molar mass (32.00 g/mol) = 0.7 mol

Since the molar ratio between HgO and O2 is 1:1, you need 0.7 moles of HgO to form 0.7 moles of O2.

Step 4: Convert moles of HgO to grams.

Since the molar mass of HgO is 216.59 g/mol, you can calculate the mass of HgO required by multiplying the number of moles by the molar mass:

mass = moles (0.7 mol) x molar mass (216.59 g/mol) = 151.61 grams

Therefore, you will need 151.61 grams of mercury(II) oxide to produce 22.4 grams of oxygen gas.