Limiting Reagent - Solution Outline

Q.
For the balanced equation shown below, if 95.3 grams of C4H9N were reacted with 220 grams of O2, how many grams of H2O would be produced?
A.
4C₄H₉N + 29O₂ => 16CO₂ + 18H₂O + 4NO₂

1. Calculate the molar masses of the three compounds:
*The molar mass of C4H9N is: (12.011)(4) + (1.00793)(9) + (14.007) = 71.183 g/mol
*The molar mass of O2 = 32.000 g/mol
*The molar mass of H2O = (1.00793)(2) + 16.000 = 18.0153 g/mol

2. Calculate the number of moles of the two reactants:
*(95.3g C4H9N)(1 mol C4H9N / 71.183 g/mol.C4H9N) = 1.339 mol C4H9N
*(220g O2)(1 mol O2 / 32.000 g O2) = 6.875 mol O2

3. Calculate the moles of C4H9N *needed* to react with 6.875 mol O2:
*(6.875 mol O2)(4 mol C4H9N / 29mol O2) = 0.9483 C4H9N [NOTE: The 4 and the 29 are coefficients from the BALANCED chemical equation]

4. *Compare* the moles of C4H9N present (1.339) to moles needed (0.9483). As you can see, we have more than the number of moles needed. That means the C4H9N is the *excess reagent* and O2 is the *limiting reagent*.

5. All the above procedure does is give us the LIMITING reagent, O2. We use the moles of the limiting reagent to get the amount of product:
*(6.875 mol O2)(18 mol H2O / 29 mol O2) = 4.27 moles H2O

6. Multiply the number of moles of H2O by the molar mass of H2O to get the grams of H2O produced.

==================