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.
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