Q5) The reaction for the combustion of propane is as follows:
C3H8 + 5O2 ---->>3CO2 + 4H2O
1. If 20.0 g of C3H8 and 20.0 g of O2 are reacted, which is the limiting reactant?
2. How many grams of H2O can be produced?
3. How many grams of the excess reactant remains at the end of the reaction?
To determine the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation. The reactant that produces fewer moles of the desired product will be the limiting reactant, as it will be completely consumed, limiting the amount of product that can be formed.
1. Let's first calculate the number of moles for each reactant.
- For C3H8 (propane) with a molar mass of 44.1 g/mol:
Number of moles = Mass / Molar mass = 20.0 g / 44.1 g/mol = 0.453 mol
- For O2 (oxygen) with a molar mass of 32.0 g/mol:
Number of moles = Mass / Molar mass = 20.0 g / 32.0 g/mol = 0.625 mol
According to the balanced equation, the stoichiometric ratio between C3H8 and O2 is 1:5. This means that for every 1 mole of C3H8, we need 5 moles of O2 to react completely.
We can calculate the mole ratio between C3H8 and O2 using the given moles:
Ratio = Moles of O2 / Moles of C3H8 = 0.625 mol / 0.453 mol ≈ 1.375
Since the mole ratio is greater than the stoichiometric ratio (1.375 > 1), it means that there are excess O2. Therefore, C3H8 is the limiting reactant as it produces fewer moles of CO2 and H2O.
2. To calculate the maximum amount of H2O that can be produced, we need to use the stoichiometric ratio between C3H8 and H2O. According to the balanced equation, the ratio is 1:4, which means for every 1 mole of C3H8, 4 moles of H2O are formed.
Number of moles of H2O formed = Moles of limiting reactant (C3H8) × Moles of H2O produced per mole of C3H8
Moles of H2O formed = 0.453 mol × 4 mol H2O/mol C3H8 = 1.812 mol
To find the mass of H2O produced, we can use the molar mass of H2O, which is 18.0 g/mol:
Mass of H2O produced = Moles of H2O formed × Molar mass of H2O = 1.812 mol × 18.0 g/mol = 32.616 g
Therefore, the maximum amount of H2O that can be produced is 32.616 grams.
3. To determine the excess reactant remaining, we need to calculate the moles of the excess reactant consumed and subtract it from the initial moles of that reactant.
Moles of O2 consumed = Moles of limiting reactant (C3H8) × Moles of O2 required per mole of C3H8 in the balanced equation
Moles of O2 consumed = 0.453 mol × 5 mol O2/mol C3H8 = 2.265 mol
Moles of O2 initially = 0.625 mol (given)
Moles of excess O2 remaining = Moles of O2 initially - Moles of O2 consumed
Moles of excess O2 remaining = 0.625 mol - 2.265 mol = -1.640 mol
Since the result is negative, it means that there is no excess O2 remaining at the end of the reaction.
Therefore, there is no excess reactant (O2) remaining at the end of the reaction.
Write the equation and balance it.
2a. Convert 20.0 g C3H8 to moles.
2b. Convert 20.0 g O2 to moles.
3a. Using the coefficients in the balanced equation, convert moles C3H8 to moles H2O.
3b. Same procedure convert moles O2 to moles H2O.
3c. You have two different answers and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller number and the reagent producing that number is the limiting reagent.
4. Using the smaller number, convert moles to grams. g = moles x molar mass.
To calculate grams excess reagent remaining, use step 3a or 3b above to convert moles of the non-limiting reagent to moles of limiting reagent, convert to grams, and subtract from 20.0.