Iron is obtained commercially by the reaction of hematite (Fe2O3) with carbon monoxide. How many grams of iron is produced when 35.0 moles of hematite react with 41.5 moles of carbon monoxide?
Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. You may want to locate the arrow keys on your computer and use them.
Fe2O3(s) + 3CO(g)--> 2Fe(s) + 3CO2(g)
mols Fe2O3 = 35.0
mols CO = 41.5
Use the coefficients in the balanced equation, convert mols Fe2o3 to mols Fe.
Do the same for mols CO to mols Fe.
It is likely that the two values will not agree so one must be wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value convert mols Fe to grams. g = mols x molar mass.
Post your work if you get stuck.
To determine the grams of iron produced when 35.0 moles of hematite react with 41.5 moles of carbon monoxide, we need to use the stoichiometry of the balanced chemical equation.
The stoichiometry tells us that the molar ratio between Fe2O3 and Fe is 2:2, which means that 1 mole of hematite produces 2 moles of iron.
Let's start calculating:
1. Calculate the number of moles of iron produced:
Number of moles of Fe = 35.0 moles of Fe2O3 x (2 moles of Fe / 2 moles of Fe2O3) = 35.0 moles of Fe
2. Calculate the molar mass of iron (Fe):
Molar mass of Fe = 55.845 g/mol
3. Calculate the mass of iron produced:
Mass of Fe = Number of moles of Fe x Molar mass of Fe
Mass of Fe = 35.0 moles x 55.845 g/mol = 1959.075 g
Therefore, approximately 1959.075 grams of iron is produced when 35.0 moles of hematite react with 41.5 moles of carbon monoxide.
To find out how many grams of iron are produced, we need to use the balanced equation and the stoichiometry of the reaction.
The balanced equation states that one mole of hematite (Fe2O3) reacts with three moles of carbon monoxide (CO) to produce two moles of iron (Fe) and three moles of carbon dioxide (CO2). This gives us the following stoichiometric ratios:
1 mol Fe2O3 : 3 mol CO : 2 mol Fe
We have 35.0 moles of Fe2O3 and 41.5 moles of CO. To determine the limiting reactant, we need to compare the number of moles of each reactant based on the stoichiometric ratio.
For Fe2O3:
35.0 moles Fe2O3 × (3 moles CO / 1 mole Fe2O3) = 105.0 moles CO
For CO:
41.5 moles CO
Since we have an excess of CO (41.5 moles) compared to Fe2O3 (105.0 moles CO), CO is the limiting reactant. This means that all the Fe2O3 will react with the available CO, and the amount of product formed will be determined by the limiting reactant.
Now, let's calculate the moles of Fe that can be produced from the limiting reactant (CO):
105.0 moles CO × (2 moles Fe / 3 moles CO) = 70.0 moles Fe
Finally, we can convert moles of Fe to grams by multiplying by the molar mass of iron (Fe):
70.0 moles Fe × (55.845 g Fe / 1 mol Fe) = 3912.15 g Fe
Therefore, when 35.0 moles of hematite react with 41.5 moles of carbon monoxide, approximately 3912.15 grams of iron are produced.