A 500g iron (Fe) rod was left in air for a long period of time, and got covered with rust (Fe2O3) as a result of reaction with atmospheric oxygen. The mass of the rusty rod was determined as 560g.
a) calculate the theoretical yield of Fe2O3(s), that is the mass of Fe2O3(s) that would form if all Fe reacted. For your convenience, molecular weights of Fe and Fe2O3 are 55.845 g/mol and 159.69 g/mol, respectively.
b)calculate the mass of Fe2O3(s) in the rusty rod?
My work:
a.) Fe= 55.845/500g= 0.11169
Fe2O3= 159.69/560=0.285
I'm stuck on what to do next.
Also I don't get how to do part b.
Please help Thanks!
there's a part c that I didn't know how to do either...
c) Calculate the actual percent yield of Fe2O3(s)
4Fe + 3O2 => 2Fe2O3
mols Fe = 500 g/atomic mass Fe = ?
Use the equation to convert mols Fe to mols Fe2O3. That will be
mols Fe x (2 mols Fe2O3/4 mol Fe) = ?mols Fe x (1/2) = ?? This is theoretical yield.
Grams Fe2O3 = mols x molar mass
%yield = (actual yield/theoretical yield)*100 = ?
To calculate the theoretical yield of Fe2O3 (mass of Fe2O3 that would form if all of the Fe reacted), we can use stoichiometry and the balanced chemical equation for the reaction between Fe and O2.
The balanced equation is:
4 Fe + 3 O2 → 2 Fe2O3
a) To calculate the theoretical yield:
1. Convert the mass of Fe to moles using the molar mass of Fe:
moles of Fe = mass of Fe / molar mass of Fe = 500 g / 55.845 g/mol ≈ 8.9492 mol
2. Use the stoichiometric ratio from the balanced equation to determine the moles of Fe2O3 that would form if all of the Fe reacted:
moles of Fe2O3 = (moles of Fe) * (2 moles of Fe2O3 / 4 moles of Fe) = 8.9492 mol * (2/4) ≈ 4.4746 mol
3. Calculate the mass of Fe2O3 using the molar mass of Fe2O3:
mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3 = 4.4746 mol * 159.69 g/mol ≈ 713.45 g
Therefore, the theoretical yield of Fe2O3 is approximately 713.45 g.
Now let's move on to part b.
b) To calculate the mass of Fe2O3 in the rusty rod:
The mass of the rusty rod is given as 560 g, which represents the combined mass of Fe and Fe2O3.
1. Subtract the mass of Fe from the total mass to find the mass of Fe2O3:
mass of Fe2O3 = mass of rusty rod - mass of Fe = 560 g - 500 g = 60 g
Therefore, the mass of Fe2O3 in the rusty rod is 60 g.