Combining 0.320 mol Fe2O3 with excess carbon produced 10.8 g Fe . Fe 2 O 3 + 3 C ⟶ 2 Fe + 3 CO a. What is the actual yield of iron in moles? b. What is the theoretical yield of iron in moles?c. What is the percent yield?

a. 0.193 moles of Fe, actual yield

b. 0.640 moles of Fe, the theoretical yield

c. 30 % yield

Explanation:

The reaction is this: Fe₂O₃  +  3C  →  2Fe  + 3CO

First of all, we determine the moles of produced Fe.

10.8 g / 55.85 g/mol = 0.193 moles → Actual yield of iron in moles

We have 0.320 moles, so we can calculate the theoretical yield by a rule of three.

1 mol of Fe₂O₃ can produce 2 moles of Fe

0.320 moles of oxide, may produce (0.320 . 2) / 1 = 0.640 moles

As we determined the theoretical yield and the actual yield, we can know the percent yield  → (Produced yield / Theoretical Yield) . 100 = % yield

(0.193 mol / 0.640 mol) . 100 = 30%

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.