Respuesta :
Answer:
1) (a) If you start with 150 g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe? Ans = 104.95g
(b) If your actual yield is 87.9 g, what is the percent yield? Ans = 83.75%
Fe2O3(s)+ 3 CO(g) ------> 2 Fe (s) + 3 CO2(g)
Explanation:
Fe = 55.8 g/mol
O = 16.0 g/mol
Molar mass of Fe2CO3 = (55.8*2) + (16.0*3)
Molar mass of Fe2CO3 = 159.6 g/mol
150g of Fe2O3 contains x moles
159.6g contains 1 mole
150*1 / 159.6 = 0.9398 moles
Number of participating moles in reaction = 2
Therefore; 2 * 0.9398 = 1.8796 moles are in the reaction.
Number of moles = mass / molar mass
Mass = number of moles * molar mass
Mass = 1.8796 * 55.84 = 104.95g of Fe.
104.95g of Fe is theoretical yield.
B. % yield = (actual yield / theoretical yield) *100
Actual yield = 87.9g
% yield = (87.9 / 104.95) * 100
% yield = 83.75%
(a) When the theoretical yield of Fe is = 104.95g
(b) = 83.75% Fe2O3(s)+ 3 CO(g) ------> 2 Fe (s) + 3 CO2(g)
What is the Theoretical yield of Fe?
A). If Fe is = 55.8 g/mol
Then O is = 16.0 g/mol
When the Molar mass of Fe2CO3 is = (55.8*2) + (16.0*3)
Then Molar mass of Fe2CO3 is = 159.6 g/mol
After that 150g of Fe2O3 contains x moles
Then 159.6g contains 1 mole
Now, 150*1 / 159.6 = 0.9398 moles
Then Number of participating moles in reaction = 2
Thus; 2 * 0.9398 = 1.8796 moles are in the reaction.
Then Number of moles is = mass / molar mass
After that Mass = number of moles * molar mass
Then Mass = 1.8796 * 55.84 = 104.95g of Fe.
Now, 104.95g of Fe is the theoretical yield.
B. Then % yield is = (actual yield / theoretical yield) *100
Now, Actual yield = 87.9g
Then % yield = (87.9 / 104.95) * 100
Therefore, % yield = 83.75%
Find more information about Theoretical yield of Fe here:
https://brainly.com/question/1492320
