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?
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.
Fe2O3 + 3C ==> 2Fe + 3CO
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? see below
C. what is the limiting reactant????
Answered in the details of step b below.
D. if you only obtained 1.25g of carbon monoxide, what is your percent yield for this reaction?
Theoretical yield is the number from part b above (which is step 4 in the details below).
%yield = [1.25g/theoretical] x 100 = ??
For (b)This is a limiting reagent problem.
Step 1. Write a balanced equation. That is done.
Step 2. Convert what you have (in this case Fe2O3 and C) to mols. Remember mols = g/molar mass
2a. mols Fe2O3=3.19/molar mass Fe2O3.
2b. mols C = 1.00g/molar mass C.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (Fe2O3 and C) to mols of what you want (in this case CO).
3a. from Fe2O3. ??mols Fe2O3 x (3 mols CO/1 mol Fe2O3)= xx mols CO.
3b. from C. ??mols C x (3 mols CO/3 mols C) = yy mols CO.
3c. Obviously, both answers can't be correct so you must determine which (either 3a or 3b) it is. The correct answer will always be the smaller of 3a or 3b. That is the limiting reagent and that will be the number used for the next step.
Step 4. Convert mols CO to grams. Just reverse step 2 above. That is mols = g/molar mass; therefore, mols x molar mass = grams.
Post your work if you get stuck. I hope this helps.
To solve this problem, we need to follow these steps:
A) Write a balanced equation for the reaction:
Fe2O3 + 3C ==> 2Fe + 3CO
B) Calculate the number of grams of carbon monoxide produced:
1. Convert the mass of iron III oxide (Fe2O3) to moles. We use the molar mass of Fe2O3 to convert grams to moles. The molar mass of Fe2O3 is 159.69 g/mol.
Moles of Fe2O3 = 3.19 g / 159.69 g/mol = 0.02 moles
2. Convert the mass of coke (C) to moles. The molar mass of C is 12.01 g/mol.
Moles of C = 1.00 g / 12.01 g/mol = 0.08 moles
3. Use the mole ratio from the balanced equation (3 moles of CO produced per 1 mole of Fe2O3) to determine the number of moles of CO produced from Fe2O3.
Moles of CO from Fe2O3 = 0.02 moles Fe2O3 * (3 moles CO / 1 mole Fe2O3) = 0.06 moles
4. Use the mole ratio from the balanced equation (3 moles of CO produced per 3 moles of C) to determine the number of moles of CO produced from C.
Moles of CO from C = 0.08 moles C * (3 moles CO / 3 moles C) = 0.08 moles
5. Compare the moles of CO produced from Fe2O3 and C. The limiting reactant is the one that produces the smaller number of moles of CO. In this case, Fe2O3 yields 0.06 moles of CO, while C yields 0.08 moles of CO. Therefore, Fe2O3 is the limiting reactant.
6. Use the molar mass of CO (28.01 g/mol) to convert the moles of CO (0.06 moles) to grams.
Grams of CO = 0.06 moles CO * 28.01 g/mol = 1.68 g
C) The limiting reactant is Fe2O3.
D) The percent yield can be calculated using the formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Given that the actual yield is 1.25 g and the theoretical yield is 1.68 g (calculated in step 6 of part B), we can plug these values into the formula:
Percent yield = (1.25 g / 1.68 g) * 100 = 74.4%
Therefore, the percent yield for this reaction is 74.4%.