In this experiment, you were asked to calculate your percent yield of alum assuming that you started with pure aluminum. Actuall
y, the aluminum had a plastic coating on it. Will the percent yield calculated (assuming pure aluminum metal) be higher, lower, or the same as the true percent yield (considering the plastic coating)? Explain
1 answer:
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Answer: There will be 131.3 tons of that will be generated.
Explanation : We have the data as 72 tons of ,
is 101 tons and
as 38 tons.
Conversion of 1 ton to kg will be approximately 907.2 Kg
We have to convert the given data of tons into kg for all;
So, moles of = (72 tons X 907.2 Kg/ton X 1000 g/Kg) / (34.082 g/ mol) = 1.916 X
moles
Now, moles of = (101 tons X 907.2 Kg/ton X 1000 g/Kg) / (32 g/mol) = 2.80 X
moles
And, = (38 tons X 907.2 Kg/ton X 1000 g/Kg) / (18.016 g/mol)
= 1.03 X moles
Now, we have to calculate the moles of
so, 1.916 X moles / 2 moles of
= 958 X
Now, for ;
So, 2.80 X moles / 3 moles of
= 933.3 X
moles
Here, we can see that is acting as a limiting reactant.
Now, moles of = 2.80 X
X (2 moles of
/ 3 moles of
) = 1.86 X
Now converting this into mass,
Mass = Moles X Molar Mass
Mass = 1.86 X X (64.066 g/mol X
kg/g) X (1 ton / 907.2 Kg)
= 131.3 tons of