How many grams of iron can be produced from 4.50Kg Fe2O3?
Fe2O3 (s) + 3CO(g)--> 2Fe (s) + 3CO2 (g)
Answer:
4500g Fe2O3 * (1 mol Fe2O3 / 159.68g Fe2O3) * (2 mol Fe / 1 mol Fe2O3) * (55.84g Fe / 1 mol Fe) = 3147g Fe
This of course assumes 100% efficiency. Hope this helps!
3.15kg
Hope it's correct...
Atomic weight of iron = 55.845amu
Atomic weight of oxygen = 15.9994amu
4500g x 2(Fe) / { 2(Fe) + 3(O) }
4500g x 2(55.845) / { 2(55.845) + 3(15.9994) }
4500g x {111.69 / 159.6882}
[ 3,147.4 grams ]
3146.6 grams of Iron.
4.50 Kg = 4500 g
Molecular weight Fe2O3 = 159.7 g/mol
4500 g / 159.7 = 28.2 moles
The ratio between Fe2O3 and Fe is 1 : 2
We would get 2 x 28.2 = 56.4 moles Fe
56.4 mol x 55.847 = 3150 g = 3.150 Kg Fe produced
no. of moles of Fe2O3
= 4500/ (2 x 55.8 + 16.0 x 3)
= 28.1955 mol.
From the equation,
reacting ratio of Fe2O3 : Fe
= 1 : 2
= 28.1955 mol : 56.3910 mol
max mass of Fe obtainable
= 56.3910 x (55.8)
= 3146.6 g
You have to first convert Kg to grams.
4.5kg x 1000 g = 4500g Fe2O3
4500g x 2molFe x 55gFe2 / 158gFe2O3 x 1molFe2 =
3132 g Fe
molar mass of Fe2O3 = 159.7 g/ mol
use farmula...no of mol = mass /molar mass
no of moles = 4500 / 159.7 = 28.18 mol
From the question 1 mol of Fe2O3 produces 2 mol of Fe. so
2 * 28.18 = 56.35 mol of Fe
Therefor...
56.35 * 55.84 = 3146 g of Fe. Ans.
(Here 55.84 = atomic mass of Fe)
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: