How many grams of calcium oxide, CaO, can be produced from 4.20 g of calcium metal and 1.60 g of oxygen gas?

Yeah Erika, stop changing names with each post smh

Well, calcium and oxygen decided to have a little chemistry party. So, calcium brought along its 4.20 g and oxygen brought 1.60 g of oxygen gas. But remember, in chemistry, it's all about the ratio.

Now, if we look at the balanced equation, we see that it's 2 moles of calcium to 2 moles of oxygen to produce 2 moles of calcium oxide.

Since we have a limited amount of calcium (4.20 g), let's find out how much oxygen we need to react completely with it. The molar mass of calcium is 40.08 g/mol, so we have 0.105 mol of calcium.

Now, let's calculate the amount of oxygen needed. The molar mass of oxygen is 16.00 g/mol, so we have 0.100 mol of oxygen.

From the balanced equation, we know that 2 moles of calcium oxide are produced for every 2 moles of calcium and 2 moles of oxygen.

So, since we have 0.100 moles of calcium oxide, we can basically say "bye-bye" to the extra calcium (0.005 mol) and oxygen (0.060 mol). They had a good time, but they were left out of the final product.

So, to answer your question, 0.100 mol of calcium oxide can be produced. But since you're asking for grams, we need to do one more calculation.

The molar mass of calcium oxide is 56.08 g/mol, so we multiply 0.100 mol by 56.08 g/mol to get 5.608 g.

Therefore, from 4.20 g of calcium metal and 1.60 g of oxygen gas, we can produce about 5.608 g of calcium oxide. That's a pretty calcium-tastic result!

To find out how many grams of calcium oxide (CaO) can be produced from the given amounts of calcium metal and oxygen gas, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be produced.

1. Start by writing the balanced chemical equation for the reaction:
2 Ca + O₂ ⟶ 2 CaO

2. Calculate the number of moles for each reactant:
Moles of calcium (Ca) = mass of calcium (g) / molar mass of calcium (g/mol)
= 4.20 g / 40.08 g/mol (molar mass of calcium)
Moles of oxygen (O₂) = mass of oxygen (g) / molar mass of oxygen (g/mol)
= 1.60 g / 32.00 g/mol (molar mass of oxygen)

3. Use the coefficients from the balanced equation to compare the number of moles of each reactant. The ratio is 2:1 for Ca to O₂, as shown in the balanced equation above.

4. Find the limiting reactant by determining which reactant has the smaller number of moles. If the ratio was not 2:1, you would need to adjust the ratio accordingly.

5. Assume calcium (Ca) is the limiting reactant. Calculate the number of moles of calcium oxide (CaO) that can be formed using the stoichiometry of the balanced equation:
Moles of CaO = Moles of limiting reactant (Ca) * (1 mol CaO / 2 mol Ca)
= Moles of Ca * (1 mol CaO / 2 mol Ca)

6. Calculate the mass of calcium oxide (CaO) that can be produced using the number of moles of CaO and the molar mass of CaO (56.08 g/mol):
Mass of CaO = Moles of CaO * molar mass of CaO

By following these steps, you should be able to find the amount of calcium oxide (CaO) that can be produced from the given amounts of reactants.

This is a limiting reagent problem.

1. Write the equation and balance it.
2Ca + O2 ==> 2CaO

2. Convert 4.20 g Ca to mols. Convert 1.60 g O2 to mols remembering that mols = grams/molar mass.

3. Using the coefficients in the balanced equation, convert mols Ca from step 2 to mols CaO. Do the same and convert mols O2 to mols CaO. Both of these numbers can't be correct. The smaller number will be the correct one to use and that will be the limiting reagent.

4. Convert mols CaO from step 3 to grams CaO. Mols x molar mass = grams.

Post your work if you get stuck.

Erika, please don't change names with each post.