A mixture of FeO and Fe2O3 with a mass of 10.0 g is converted to 7.48 g of pure Fe metal. What is the amount of FeO and Fe2O3 (in grams) in the original sample.
Not sure where to start. Any help is appreciated!
mm = molar mass
am = atomic mass
You have two equations and you solve the two equations simultaneously.
Let X = grams FeO and
let Y = grams Fe2O3.
equation 1 is X + Y = 10.0 g
The second equation is converting this 10.0 g into pure iron in terms of X and Y.
eqn 2 is X(am Fe/mm FeO) + Y(2 am Fe/mm Fe2O3) = 7.48 g
Solve these two equation for X and Y giving you grams FeO and Fe2O3.
Post your work if you get stuck. This is the chem part. The only thing left is the math.
To solve this problem, we can set up a system of equations using the given information. Let's denote the mass of FeO as x and the mass of Fe2O3 as y.
1. Write the equation for the total mass of the original sample:
x + y = 10.0 g -- Equation 1
2. Write the equation for the mass of pure Fe produced:
FeO: (x / 72.9 g/mol) x (55.8 g/mol) = (55.8/72.9) x -- Equation 2
Fe2O3: (y / 159.7 g/mol) x (55.8 g/mol) = (55.8/159.7) y -- Equation 3
3. Write the equation for the total mass of Fe produced:
(55.8/72.9) x + (55.8/159.7) y = 7.48 g -- Equation 4
Now we have a system of three equations with three unknowns (x, y, and the total mass of Fe produced).
To solve this system of equations, we can use substitution or elimination methods. Let's solve it using the substitution method:
From Equation 1, we can express x in terms of y:
x = 10.0 g - y
Substitute this expression into Equation 4:
(55.8/72.9)(10.0 g - y) + (55.8/159.7) y = 7.48 g
Now we can solve this equation for y:
(55.8/72.9)(10.0 g) - (55.8/72.9) y + (55.8/159.7) y = 7.48 g
5580/72.9 - (55.8/72.9 - 55.8/159.7) y = 748/100
76.4824 - (0.766740 - 0.34969) y = 7.48
76.4824 - 0.766740y + 0.34969y = 7.48
76.4824 - 0.41705y = 7.48
-0.41705y = 7.48 - 76.4824
-0.41705y = -69.0024
y = (-69.0024) / (-0.41705)
y ≈ 165.493 g
Now substitute the value of y back into Equation 1 to find x:
x + 165.493 g = 10.0 g
x = 10.0 g - 165.493 g
x ≈ -155.493 g
Since the mass cannot be negative, we conclude that there is no FeO in the original sample. The mass of Fe2O3 in the original sample is approximately 165.493 grams.