Calculate the number of grams of CO2 produced by the reaction of 2.45 g of iron III oxide and 1.12 g C

Question

2Fe2O3+3C—>4Fe+3CO2

partial answer is 1.XX

Answer 1

So first we have to identify the limiting reactant in the problem. We can do this by converting the amount of each reactant into moles. Divide the mass available of each reactant by their respective molar masses.

Fe₂O₃: 2.45 grams ÷ 159.68 grams/mol = 0.0153 moles

C: 1.12 grams ÷ 12.011 grams/mol = 0.0932 moles

Since we need 3 moles of carbon for every 2 moles of iron oxide in this reaction, and we have almost 9x more carbon, iron oxide is our limiting reactant.

Now for some stoichiometry to calculate the grams of CO₂ produced.

[2.45 grams Fe₂O₃ ] [1 mole Fe₂O₃ ] [ 3 moles CO₂] [44.01 grams/mole CO₂] = 1.01 grams of CO₂

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ [159.68 grams/mol] [2 moles Fe₂O₃ ]¯¯[1 mole CO₂]¯¯¯¯¯

Hope this helps!

Answer 2

For this problem, you will want to find out whether the limiting reactant is the 2.45g of Fe₂O₃ or the 1.12g of carbon (C).

Set up a stoichiometric table to see how much of which reactant will yield how much CO₂ (IMPORTANT: the limiting reactant yields the lower mass of product).

Find the molar mass of Fe₂O₃ and that of C. Use the molar masses and the coefficients from the equation to solve for how many grams of CO₂ are produced given each scenario.

The amount of CO₂ produced by Fe₂O₃:

2.45g Fe₂O₃ 1 mol Fe₂O₃ 3 mol CO₂ 44.01g CO₂

––––––––––– x –––––––––– x –––––––––– x –––––––––– = ?g CO₂

1 159.69g Fe₂O₃ 2 mol Fe₂O_{3 1}

The amount of CO₂ produced by C:

1.12g C 1 mol C 3 mol CO₂ 44.01g CO₂

––––––––––– x –––––––––– x –––––––––– x –––––––––– = ??g CO₂

1 12.01g C 3 mol C 1 mol CO₂

Solve for each mass of Fe₂O₃ by plugging these equations into your calculator. Whichever mass is the lowest is your answer.

Does any of this not make sense?