if 100 g of iron were reacted with 100 g of oxygen according to the reaction below, the maximum yield in grams of Fe2O3 would be:

Question

if 100 g of iron were reacted with 100 g of oxygen according to the reaction below, the maximum yield in grams of Fe2O3 would be:

Dr. Bob. Please help.

Answer 1

Balance the equation. Note the coefficents on Fe and O2, that is the mole ratio.

calcuate how many moles of Fe, and O2 you have.

Which is the limiting reactant.

For instance, if the coefficents were 45Fe+ 34O2, the mole ratio of O2 to Fe is 34/45

If you had 2.3 moles of Fe, and 3.2 moles of O2, your mole ratio of O2 to Fe would be 3.2/2.3, significantly greater than the coefficents of needed moles. Well, if you have more, that means you have too much O2, so you limiting reageant would be Fe.

Finally, figure out how much product you get with your limiting reageant.

Answer 2

This is a limiting reagent problem. How do I know? Because BOTH reactants are given.

You apparently have the reaction in the problem but didn't post it. I assume it is as follows: Check it to see!
These actually are stoichiometry problems but being limiting reagent we must work them twice and take the smaller product.

4Fe + 3O2 ==> 2Fe2O3

Convert 100 g Fe to moles. moles = grams/molar mass.
Convert 100 g oxygen to moles. Same method.

Using the coefficients in the balanced equation, convert moles Fe to moles of the product.
Same process convert moles O2 to moles of the procuct.
It is most likely that the two answers for moles product formed will not be the same which means one of them is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.

Now convert the value from the last step to grams. g = moles x molar mass.

Answer 3

To determine the maximum yield of Fe2O3 (iron oxide), we first need to find the limiting reactant. The limiting reactant is the one that is completely consumed during the reaction, thus determining the maximum amount of product that can be formed.

Let's first write out the balanced chemical equation for the reaction:

4 Fe + 3 O2 -> 2 Fe2O3

From the balanced equation, it is evident that the mole ratio between Fe and Fe2O3 is 4:2, or simply 2:1. This means that for every 2 moles of Fe, we will get 1 mole of Fe2O3.

Now we need to find the number of moles of Fe and O2 present in the reaction. We can do this using the molar mass of each element.

Molar mass of Fe (iron) = 55.845 g/mol
Molar mass of O2 (oxygen) = 32.00 g/mol (each O atom has a molar mass of 16.00 g/mol)

Number of moles (n) = mass (m) / molar mass (M)

For Fe:
n(Fe) = 100 g / 55.845 g/mol

For O2:
n(O2) = 100 g / 32.00 g/mol

Now we need to compare the number of moles of Fe and O2 to determine which is the limiting reactant. Since the mole ratio between Fe and Fe2O3 is 2:1, the number of moles of Fe2O3 that can be formed will be half the number of moles of Fe.

n(Fe2O3) = n(Fe) / 2

Now, to find the maximum yield of Fe2O3 in grams, we can use the number of moles and the molar mass of Fe2O3:

Mass(Fe2O3) = n(Fe2O3) * molar mass(Fe2O3)

Substituting the values we have:

Mass(Fe2O3) = (n(Fe) / 2) * [2 * (molar mass(Fe) + 3 * molar mass(O))]

Mass(Fe2O3) = (100 g / 55.845 g/mol) / 2 * [2 * (55.845 g/mol + 3 * 16.00 g/mol)]

Evaluating this expression will give you the maximum yield of Fe2O3 in grams.

Note: The above calculation assumes that the reaction goes to completion with no other factors affecting the yield. In reality, actual yields may be lower due to factors like side reactions, incomplete reactions, or impurities.