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Given this equation:
How many moles of iron (III) oxide can be produced from :
4 mol of Iron and 2.5 mol of oxygen?
How would I solve? I am learning stoichiometry overnight since I have been absent and want to be sure I am doing this right.
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`4Fe+3O_2 rarr 2Fe_2O_3`
To get 2 moles of `Fe_2O_3` we need 4 moles of Fe and 3 moles of `O_2` .
In the given mixture we have 4 moles of Fe and 2.5 moles of `O_2` .
Since we need 3 moles of `O_2` and 4 moles of Fe to get 2 moles of `Fe_2O_3` , the above reaction will not completely takes place in the mixture to form 2 moles of `Fe_2O_3` .
So what will happen is the reaction will happen until all the `O_2` is reacted.
So finally 2.5 moles of `O_2` will react.
`O_2:Fe_2O_3 = 3:2`
Amount of `Fe_2O_3` formed `= 2/3xx2.5 = 1.667`
So 1.667 moles of `Fe_2O_3` will form in the mixture.
This is a typical example of the concept of limiting reagent in stoichiometry (branch of chemistry dealing with relative amounts of reactants and products). Suppose 3 moles of A reacts with 2 moles of B to produce 2 moles of the product C. Now if 2 moles of both A and B are taken, the limiting reagent can be found out in the following way: calculate the number of moles of products that is obtainable from each of these reactants individually (assuming the other reagent to be present in sufficient amount). The reactant that produces the lesser amount of product is the ‘limiting reactant’. Here 2 moles of A should produce 2*2/3=1.33 moles of C, whereas 2 moles of B is expected to produce 2 moles of C. Thus A should be the limiting reagent, i.e. the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.
In the given balanced equation 4Fe+3O2-->2Fe2O3
4 moles of Fe reacts with 3 moles of O2 to produce 3 moles of Fe2O3. Reagents actually taken are 4 moles of Iron and 2.5 moles of oxygen. Applying the same concept to find the limiting reagent, 4 moles of Fe (in presence of sufficient oxygen) should produce 2 moles of Fe2O3, whereas 2.5 moles of O2 (in presence of sufficient iron) is expected to produce 2*2.5/3 = 1.667 moles of iron (III) oxide. Thus O2 should be the limiting reagent. Number of moles of iron (III) oxide that can be produced is thus 1.667. Next, to find the amount of excess reactant (iron), we must calculate how much of the non-limiting reactant (iron) actually did react with the limiting reactant (oxygen).
4.00 moles (amount actually taken) – 2*1.667 = 3.333 moles (amount reacted) = 0.667 moles (amount remaining excess).
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