How many moles of Fe3O4 are required to supply enough iron to prepare 0.785 mol Fe2O3?
A good short cut to remember. Just convert using the "chemical factor." That isn't taught in many schools now (perhaps even none) but it's just the ratio of the molar masses. It will convert grams to grams or moles to moles. Note I multiplied by 2 and 3 to keep the Fe atoms equal. I can show you the long way if you want to see it.
0.785 mol Fe2O3 x (2*molar mass Fe3O4/3*molar mass Fe2O3) = ?
To find out how many moles of Fe3O4 are required to supply enough iron to prepare 0.785 mol Fe2O3, we need to use the balanced chemical equation between Fe3O4 and Fe2O3.
The balanced equation for the reaction is:
Fe3O4 + 1.5 O2 ā 3 Fe2O3
From the equation, we can see that 1 mole of Fe3O4 produces 3 moles of Fe2O3. This means that the stoichiometric ratio between Fe3O4 and Fe2O3 is 1:3.
To calculate the number of moles of Fe3O4 required, we can use the following proportion:
(0.785 mol Fe2O3 / 3 mol Fe2O3) * 1 mol Fe3O4 = 0.2617 mol Fe3O4
Therefore, approximately 0.262 moles of Fe3O4 are required to supply enough iron to prepare 0.785 mol Fe2O3.
To determine the number of moles of Fe3O4 required to supply enough iron to prepare 0.785 mol of Fe2O3, we need to first establish the balanced equation for the reaction between Fe3O4 and Fe2O3.
The balanced equation for the reaction can be written as follows:
Fe3O4 + Fe2O3 -> 3Fe2O3
From the balanced equation, we can see that every one mole of Fe3O4 reacts to produce three moles of Fe2O3.
Therefore, since we want to prepare 0.785 mol of Fe2O3, we would need:
0.785 mol Fe2O3 x (1 mol Fe3O4 / 3 mol Fe2O3) = 0.262 mol Fe3O4
So, 0.262 moles of Fe3O4 would be required to supply enough iron to prepare 0.785 mol of Fe2O3.