What mass iron(III) oxide could be produced from 8.2 mol of oxygen gas in the synthesis of iron(III)?(Balanced equation)
4Fe + 3O2 ==> 2Fe2O3
Using the coefficients in the balanced equation, convert 8.2 moles oxygen to moles Fe2O3.
Now convert moles Fe2O3 to grams. grams = moles x molar mass.
To determine the mass of iron(III) oxide produced, we need the balanced chemical equation for the synthesis of iron(III) oxide.
The balanced equation for the synthesis of iron(III) oxide is:
4Fe(s) + 3O2(g) -> 2Fe2O3(s)
From the balanced equation, we can see that 4 moles of iron react with 3 moles of oxygen gas to produce 2 moles of iron(III) oxide.
Given that we have 8.2 mol of oxygen gas, we can use this information to find the number of moles of iron(III) oxide produced.
According to the stoichiometry of the balanced equation, the mole ratio of oxygen gas to iron(III) oxide is 3:2. We can set up a proportion to find the number of moles of iron(III) oxide:
(3 mol O2 / 2 mol Fe2O3) = (8.2 mol O2 / x)
Cross-multiplying, we get:
3x = 2 * 8.2
Simplifying, we have:
3x = 16.4
Dividing by 3, we find:
x ≈ 5.47 mol Fe2O3
Now that we have the number of moles of iron(III) oxide (approximately 5.47 mol), we can use the molar mass of iron(III) oxide to calculate its mass.
The molar mass of Fe2O3 is calculated as follows:
(2 x atomic mass of Fe) + (3 x atomic mass of O)
(2 x 55.845 g/mol) + (3 x 16.00 g/mol) = 55.845 g/mol + 48.00 g/mol = 103.845 g/mol
Finally, we can calculate the mass of iron(III) oxide:
Mass = number of moles x molar mass
Mass = 5.47 mol x 103.845 g/mol ≈ 566.95 g
Therefore, approximately 566.95 grams of iron(III) oxide could be produced from 8.2 moles of oxygen gas in the synthesis of iron(III).