2Hg+O2=2HgO

if 50.0 grams of mercury reacts with 50.0 grams of oxygen gas, how many grams of mercury(ii)oxide are produced?

This is a limiting reagent (LR) problem. I know that because amounts are given for BOTH reactants.

1. Convert grams Hg to mols. mols = grams/molar mass.

2. Do the same for g oxygen to mols.

3a. Using the coefficients in the balanced equation, convert mols Hg to mols HgO.
3b. Do the same to convert mols O2 to mols HgO
3c. It is likely that the two values from 3a and 3b will not agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.

4. Now use the smaller value mols HgO and convert to grams. g = mols x molar mass

NOTE: The correct way to show the name is mercury(II) oxide.(not ii)

To find out how many grams of mercury(II) oxide (HgO) are produced when 50.0 grams of mercury (Hg) reacts with 50.0 grams of oxygen gas (O2), we need to use the concept of stoichiometry. Stoichiometry allows us to determine the quantitative relationship between reactants and products in a chemical reaction.

1. Start by writing the balanced chemical equation:
2Hg + O2 → 2HgO

2. Find the molar masses of the substances involved:
- Molar mass of Hg = 200.59 g/mol
- Molar mass of O2 = 32.00 g/mol

3. Calculate the number of moles of each substance:
- Moles of Hg = mass / molar mass = 50.0 g / 200.59 g/mol = 0.2493 mol
- Moles of O2 = mass / molar mass = 50.0 g / 32.00 g/mol = 1.5625 mol

4. Determine the stoichiometric ratio between Hg and HgO. From the balanced equation, we see that 2 moles of Hg react to produce 2 moles of HgO.

5. Now we can calculate the number of moles of HgO produced:
- Moles of HgO = 2 * (moles of Hg) = 2 * 0.2493 mol = 0.4986 mol

6. Finally, calculate the mass of HgO produced using its molar mass:
- Mass of HgO = moles of HgO * molar mass of HgO = 0.4986 mol * 216.61 g/mol = 107.92 g

Therefore, when 50.0 grams of mercury reacts with 50.0 grams of oxygen gas, 107.92 grams of mercury(II) oxide are produced.