For the following reaction:
Fe2O3 + 3CO --> 2Fe + 3CO2
how many grams of Fe2O3 is needed to produce 111 g of Fe?
A) 79.3 g Fe2O3
B) 159 g Fe2O3
C) 317 g Fe2O3
D) 71.4 g Fe2O3
? g = 111 g Fe * 1 mol Fe2O3/2 mol Fe * 159.6 g/1 mol Fe2O3
= 8857.8 g of Fe2O3
answer is B.
111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 )
111g Fe * (1 mol Fe / 56g Fe ) * (1 mol Fe2O3 / 1 mol Fe) * (160g Fe2O3 / mol Fe2O3 )
= 159g Fe203
To determine the grams of Fe2O3 needed to produce 111 g of Fe, we can use the balanced chemical equation.
According to the equation, the ratio of Fe2O3 to Fe is 1:2. This means that for every 1 mole of Fe2O3, we get 2 moles of Fe.
First, we need to calculate the number of moles of Fe. We can use the molar mass of Fe (55.85 g/mol) to convert the given mass to moles.
Number of moles of Fe = Mass of Fe / Molar mass of Fe
Number of moles of Fe = 111 g / 55.85 g/mol
Number of moles of Fe = 1.98 mol
Since the ratio of Fe2O3 to Fe is 1:2, we can conclude that the number of moles of Fe2O3 needed is half of the number of moles of Fe.
Number of moles of Fe2O3 = Number of moles of Fe / 2
Number of moles of Fe2O3 = 1.98 mol / 2
Number of moles of Fe2O3 = 0.99 mol
Finally, we can convert moles of Fe2O3 to grams using its molar mass (159.69 g/mol).
Grams of Fe2O3 = Number of moles of Fe2O3 × Molar mass of Fe2O3
Grams of Fe2O3 = 0.99 mol × 159.69 g/mol
Grams of Fe2O3 = 157.72 g
Therefore, the answer is approximately 157.72 g Fe2O3. Thus, the closest option is C) 317 g Fe2O3.