Given the equation: 4 Fe + 3 O2 -> 2 Fe2O3

- If I react 5.75 g iron with 3.00 g oxygen, what is the LR?
- What is the theoretical yield of iron(III) oxide?

1 answer

To do limiting reagent problems you just work two regular stoichiometry problems, which Im sure you know how to do, then comare the products. The SMALLER number wins, like this.
4Fe + 3O2 -> 2Fe2O3

mols Fe = grams/atomic mass =5.75/55.85 = approx 0.1 but you need a more exact number.
Convert this to mols of Fe2O3 that could be produced. mols Fe2O3 produced will be 1/2 mols Fe = approx 0.05.

mols O2 = g/molar mass = 3.00/32 = approx 0.09. Again you need a better number.
Convert to mols Fe2O3 that could be produced. That will be approx 0.09 x 2/3 = approx 0.06 mols Fe2O3.
The smaller number is that of Fe; therefore, Fe is the LR and there will be an excess of O2 Now, ignoring the O2, just calculate how much Fe2O3 will be produced from the 5.75 g Fe. That will be 0.05 mols. Convert to grams by g = mols x molar mass and that is the theoretical yield.

Post your work if you gt stuck.
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