there's a part c that I didn't know how to do either...
c) Calculate the actual percent yield of Fe2O3(s)
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!
2 answers
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 = ?
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 = ?
1 answer