How many grams of O2 are needed to produce 45.8 grams of Fe2O3 in the following reaction? 4Fe(s)+3O2(g) - 2Fe2O3(s)

Back track,

45.8 grams of Fe2O3*(1 mole/159.69 g)= moles of Fe2O3

moles of Fe2O3 *(3 moles of O2/2 moles of Fe2O3)=moles of O2

moles of O2*(32.0g of O2/1mole)= mass of O2

Well, let me do some calculations with my clown calculator. According to the equation, we need 3 moles of O2 for every 2 moles of Fe2O3.

The molar mass of O2 is approximately 32 grams per mole, and the molar mass of Fe2O3 is about 160 grams per mole.

So let's calculate the moles of Fe2O3 we have: 45.8 grams of Fe2O3 divided by 160 grams per mole gives us about 0.29 moles of Fe2O3.

Now, using the mole ratio from the equation, we can determine how many moles of O2 we need. Since the ratio is 3:2, we need 3/2 times 0.29 moles of O2.

Doing the math, we find that we need approximately 0.435 moles of O2.

Converting moles to grams, 0.435 moles of O2 is equal to approximately 13.92 grams of O2.

So, to produce 45.8 grams of Fe2O3, we need approximately 13.92 grams of O2. Just try not to breathe it all in, or you might float away like a balloon!

To find the number of grams of O2 needed to produce 45.8 grams of Fe2O3, we need to use the stoichiometry of the reaction.

The stoichiometry of the reaction tells us that for every 4 moles of Fe (iron) reacting, we need 3 moles of O2 (oxygen) to produce 2 moles of Fe2O3 (iron (III) oxide).

First, we need to find the molar mass of Fe2O3:
- The atomic mass of iron (Fe) is 55.845 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
Therefore, the molar mass of Fe2O3 is (2 * 55.845) + (3 * 16.00) = 159.69 g/mol.

Next, we calculate the moles of Fe2O3:
Moles = Mass / Molar mass
Moles of Fe2O3 = 45.8 g / 159.69 g/mol ≈ 0.287 mol

Now, we can use the stoichiometry to find the moles of O2 needed:
Moles of O2 = Moles of Fe2O3 * (3 mol O2 / 2 mol Fe2O3)
Moles of O2 = 0.287 mol * (3 mol O2 / 2 mol Fe2O3) ≈ 0.431 mol

Finally, we can convert the moles of O2 to grams:
Grams = Moles * Molar mass
Grams of O2 = 0.431 mol * 32.00 g/mol (molar mass of O2) ≈ 13.79 g

Therefore, approximately 13.79 grams of O2 are needed to produce 45.8 grams of Fe2O3 in the given reaction.

9.933 grams