Iron ore consists mainly of iron III oxide. When iron III oxide is heated with an excess of coke(carbon), iron metal and carbon monoxide are produced. Assume that even if an excess of coke is not present that the reaction proceeds as described.
A)write a balanced equation for the above reaction.
B) If you started with 3.19g of iron III oxide and 1.00g of coke, how many grams of carbon monoxide would you theoretically obtain?
C. what is the limiting reactant????
D. if you only obtained 1.25g of carbon monoxide, what is your percent yield for this reaction?
Please do not double post. I have answered your question below.
A) The balanced equation for the reaction described is:
2 Fe2O3 + 3 C → 4 Fe + 3 CO
B) To determine the grams of carbon monoxide produced, we need to use stoichiometry.
We start with the given mass of iron III oxide (Fe2O3) and convert it to moles using the molar mass of Fe2O3 (159.69 g/mol).
3.19 g Fe2O3 × (1 mol Fe2O3 / 159.69 g Fe2O3) = 0.02 mol Fe2O3
Next, we compare the coefficients in the balanced equation to find the mole ratio between Fe2O3 and CO. From the balanced equation, we see that the ratio is 3:3 or 1:1.
So, the moles of carbon monoxide produced is also 0.02 mol.
Finally, we convert the moles of carbon monoxide to grams using the molar mass of CO (28.01 g/mol).
0.02 mol CO × (28.01 g CO / 1 mol CO) = 0.56 g CO
Therefore, theoretically, you would obtain 0.56 grams of carbon monoxide.
C) To determine the limiting reactant, we need to compare the number of moles of each reactant in the balanced equation with the given masses of both reactants.
First, we calculate the number of moles of coke (C) using its molar mass (12.01 g/mol).
1.00 g C × (1 mol C / 12.01 g C) = 0.083 mol C
From the balanced equation, we see that the mole ratio between Fe2O3 and C is 2:3.
Using this ratio, we can calculate the number of moles of Fe2O3 that would react with 0.083 mol of C.
0.083 mol C × (2 mol Fe2O3 / 3 mol C) = 0.055 mol Fe2O3
Based on this calculation, we can see that 0.055 mol of Fe2O3 would react with all of the 0.083 mol of C. However, we only have 0.02 mol of Fe2O3 available.
Since the actual moles of Fe2O3 (0.02 mol) is less than the calculated moles of Fe2O3 that would react with all of the C (0.055 mol), Fe2O3 is the limiting reactant.
D) Theoretical yield is the maximum amount of product that can be obtained according to the balanced equation. In this case, the theoretical yield of carbon monoxide is 0.56 g.
To calculate the percent yield, we need to compare the actual yield (1.25 g) to the theoretical yield (0.56 g) and express it as a percentage.
Percent yield = (actual yield / theoretical yield) × 100
= (1.25 g / 0.56 g) × 100
≈ 223.2%
Therefore, the percent yield for this reaction is approximately 223.2%.