The rusting of iron is represented by the equation 4Fe + 3O2 2Fe2O3. If you have a 1.75-mol sample of iron, how many moles of Fe2O3 will there be after the iron has rusted completely?
To find the number of moles of Fe2O3 after the iron has rusted completely, we need to use stoichiometry.
The balanced chemical equation for the rusting of iron is:
4Fe + 3O2 -> 2Fe2O3
From the equation, we can see that 4 moles of iron (Fe) reacts to produce 2 moles of iron(III) oxide (Fe2O3).
Given that we have a 1.75-mol sample of iron, we can set up a proportion using the stoichiometric ratio:
4 moles of Fe / 2 moles of Fe2O3 = 1.75 moles of Fe / x moles of Fe2O3
Cross-multiplying, we have:
4x = 2 * 1.75
4x = 3.5
Dividing both sides by 4, we get:
x = 3.5 / 4
x = 0.875
Therefore, there will be 0.875 moles of Fe2O3 after the iron has rusted completely.
According to the balanced equation, the ratio of Fe to Fe2O3 is 4:2, which simplifies to 2:1. This means that for every 2 moles of Fe that rust, 1 mole of Fe2O3 is produced.
Since we have a 1.75-mol sample of Fe, we can divide it by 2 to find the number of moles of Fe2O3 that will be produced.
1.75 moles of Fe / 2 = 0.875 moles of Fe2O3.
Therefore, there will be 0.875 moles of Fe2O3 produced after the iron has rusted completely.
The coefficients tell you.
4 mols Fe = 2 mols Fe2O3.
1.75 x (2/4) = ?