The dusting of iron is represented by the equation 4fe3 +3o2 —2fe2o3. if you have 1.8 mol sample of iron how many moles of fe2o3 will there be when the iron has completely rusted
To find the number of moles of Fe2O3 produced when 1.8 mol of Fe has completely rusted, we need to use the stoichiometry of the balanced equation:
4Fe + 3O2 → 2Fe2O3
From the balanced equation, we can see that for every 4 moles of Fe that react, 2 moles of Fe2O3 are produced.
Therefore, we can set up a proportion to find the number of moles of Fe2O3:
4 mol Fe / 2 mol Fe2O3 = 1.8 mol Fe / x mol Fe2O3
Cross-multiplying, we have:
4 mol Fe * x mol Fe2O3 = 1.8 mol Fe * 2 mol Fe2O3
Simplifying, we get:
4x = 3.6
Dividing both sides by 4, we find:
x = 0.9 mol Fe2O3
Therefore, when 1.8 mol of Fe has completely rusted, there will be 0.9 mol of Fe2O3 produced.
To find the number of moles of Fe2O3 produced when 1.8 moles of Fe react completely, we can use the stoichiometry represented by the balanced equation:
4Fe + 3O2 → 2Fe2O3
According to the balanced equation, for every 4 moles of Fe, we get 2 moles of Fe2O3. So, we can set up a proportion to find the number of moles of Fe2O3.
(4 moles Fe / 2 moles Fe2O3) = (1.8 moles Fe / x moles Fe2O3)
Cross-multiplying gives us:
4 moles Fe * x moles Fe2O3 = 2 moles Fe2O3 * 1.8 moles Fe
Therefore, x = (2 moles Fe2O3 * 1.8 moles Fe) / 4 moles Fe
Simplifying the expression, we get:
x = 0.9 moles Fe2O3
Therefore, when 1.8 moles of Fe have completely rusted, there will be 0.9 moles of Fe2O3 produced.