Consider the following reaction.

4Fe + 3O2=2Fe2O3

a. How many moles of O2 are required to react completely with 29.8 mg of Fe?

4.00 x 10^-4 moles O2

b. How many grams Fe2O3, are produced from the complete reaction of 17.8 g of Fe?

25.4 g Fe2O3

right. So why did you have a problem with the earlier post?

To solve this reaction stoichiometry problem, we need to use the balanced equation given:

4Fe + 3O2 = 2Fe2O3

a. To find the number of moles of O2 required to react completely with 29.8 mg of Fe, we first need to convert the mass of Fe to moles. Here's how to calculate it step by step:

Step 1: Convert the mass of Fe to grams:
29.8 mg * (1 g / 1000 mg) = 0.0298 g Fe

Step 2: Convert the mass of Fe to moles using its molar mass:
0.0298 g Fe * (1 mol Fe / 55.85 g Fe) = 5.32 x 10^-4 moles Fe

Step 3: Determine the mole ratio between O2 and Fe from the balanced equation:
4 moles Fe : 3 moles O2

Step 4: Calculate the moles of O2 required:
5.32 x 10^-4 moles Fe * (3 moles O2 / 4 moles Fe) = 4.00 x 10^-4 moles O2

Therefore, 4.00 x 10^-4 moles of O2 are required to react completely with 29.8 mg of Fe.

b. To find the mass of Fe2O3 produced from the complete reaction of 17.8 g of Fe, we'll follow these steps:

Step 1: Convert the mass of Fe to moles using its molar mass:
17.8 g Fe * (1 mol Fe / 55.85 g Fe) = 0.318 mol Fe

Step 2: Determine the mole ratio between Fe2O3 and Fe from the balanced equation:
4 moles Fe : 2 moles Fe2O3

Step 3: Calculate the moles of Fe2O3 produced:
0.318 mol Fe * (2 mol Fe2O3 / 4 mol Fe) = 0.159 mol Fe2O3

Step 4: Convert the moles of Fe2O3 to grams using its molar mass:
0.159 mol Fe2O3 * (159.69 g Fe2O3 / 1 mol Fe2O3) = 25.4 g Fe2O3

Therefore, 25.4 g of Fe2O3 are produced from the complete reaction of 17.8 g of Fe.

To find the number of moles of O2 required to react completely with 29.8 mg of Fe, you need to use the balanced equation for the reaction:

4Fe + 3O2 → 2Fe2O3

First, convert the mass of Fe from milligrams to grams:

29.8 mg = 0.0298 g

Next, calculate the number of moles of Fe using its molar mass:

Molar Mass of Fe = 55.845 g/mol

moles of Fe = mass of Fe / molar mass of Fe
= 0.0298 g / 55.845 g/mol
= 0.000532 mol

According to the balanced equation, the molar ratio between Fe and O2 is 4:3. Therefore, you can use this ratio to find the number of moles of O2:

moles of O2 = (moles of Fe) x (3 moles of O2 / 4 moles of Fe)
= 0.000532 mol x (3/4)
= 0.000399 mol

Therefore, the number of moles of O2 required to react completely with 29.8 mg of Fe is approximately 0.000399 mol or 4.00 x 10^-4 mol (rounded to four significant figures).

To find the number of grams of Fe2O3 produced from the complete reaction of 17.8 g of Fe, you can use the balanced equation and molar masses:

4Fe + 3O2 → 2Fe2O3

Molar Mass of Fe2O3 = 159.6882 g/mol

First, calculate the number of moles of Fe using its molar mass:

moles of Fe = mass of Fe / molar mass of Fe
= 17.8 g / 55.845 g/mol
= 0.3185 mol

Next, use the molar ratio between Fe and Fe2O3 (2:4) to find the number of moles of Fe2O3:

moles of Fe2O3 = (moles of Fe) x (2 moles of Fe2O3 / 4 moles of Fe)
= 0.3185 mol x (2/4)
= 0.1593 mol

Finally, calculate the mass of Fe2O3 using its molar mass:

mass of Fe2O3 = moles of Fe2O3 x molar mass of Fe2O3
= 0.1593 mol x 159.6882 g/mol
= 25.4 g

Therefore, the mass of Fe2O3 produced from the complete reaction of 17.8 g of Fe is 25.4 g.