Iron is obtained from iron ore according to the following reaction: Fe2O3+CO---> Fe+CO2 (unbalanced).

a. Assuming the blast furnace is 90.0% efficient in recovering the iron, what is the actual mass of iron obtainable from a ton of ore?
b. What is theoretical yield of iron if 200.0g of Fe2O3 reacts with 100.0g of CO?
c. How many grams of excess reactant are left at the end of the reaction?

Here is a solved example of a stoichiometry problem. It will work a.

For b, this is a limiting regent problem. I work this type by doing two of the simple stoichiometry problems and using the reagent that produces the smallest value for moles of the product.
c is same type as a.

http://www.jiskha.com/science/chemistry/stoichiometry.html

Thank you.

To solve these problems, we need to use stoichiometry and the concept of limiting reactant. Let's break it down step by step:

a. To find the actual mass of iron obtainable from a ton of ore, we first need to calculate the moles of iron found in the iron ore.

1. Calculate the molar mass of Fe2O3 (iron(III) oxide):
Molar mass(Fe2O3) = 2 * Molar mass(Fe) + 3 * Molar mass(O)
Molar mass(Fe2O3) = 2 * 55.85 g/mol + 3 * 16.00 g/mol
Molar mass(Fe2O3) = 159.70 g/mol

2. Convert the mass of the ore (1 ton = 1000 kg) to moles:
Moles(Fe2O3) = Mass(Fe2O3) / Molar mass(Fe2O3)
Moles(Fe2O3) = 1000 kg / 159.70 g/mol

3. Convert the moles of Fe2O3 to moles of Fe using the balanced equation:
According to the balanced equation, 1 mol of Fe2O3 produces 2 mol of Fe.
Moles(Fe) = Moles(Fe2O3) * 2

4. Calculate the actual mass of iron obtainable, taking into account the 90.0% efficiency:
Actual mass(Fe) = Moles(Fe) * Molar mass(Fe) * Efficiency
Actual mass(Fe) = Moles(Fe) * 55.85 g/mol * 0.90

b. To find the theoretical yield of iron, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thereby limiting the amount of product formed.

1. Calculate the moles of each reactant:
Moles(Fe2O3) = Mass(Fe2O3) / Molar mass(Fe2O3)
Moles(CO) = Mass(CO) / Molar mass(CO)

2. Determine the moles of Fe produced based on the balanced equation:
According to the balanced equation, 1 mol of Fe2O3 produces 1 mol of Fe.
Moles(Fe) = Moles(Fe2O3)

3. Compare the moles of Fe obtained from the reactants (Fe2O3 and CO) and identify the limiting reactant:
The limiting reactant is the reactant that produces fewer moles of Fe.
Theoretical yield(Fe) = Moles(Fe) * Molar mass(Fe)

c. To find the grams of excess reactant left at the end of the reaction, we need to determine the amount of excess reactant used and subtract it from the initial amount.

1. Determine the moles of Fe produced based on the limiting reactant:
Moles(Fe) = Moles(Fe2O3) (the same as in part b)

2. Calculate the amount of CO used based on the balanced equation:
According to the balanced equation, 1 mol of Fe2O3 reacts with 1 mol of CO.
Moles(CO used) = Moles(Fe2O3)

3. Calculate the amount of CO left:
Moles(CO left) = Moles(CO initial) - Moles(CO used)

4. Calculate the mass of CO left:
Mass(CO left) = Moles(CO left) * Molar mass(CO)

By following these steps, you should be able to solve all three parts of the problem.