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.
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
1 answer