Summary
Highlights
The video introduces the concepts of limiting reactants, excess reactants, theoretical yield, and percent yield. It begins with an example combustion reaction of propane (C3H8) reacting with oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O), emphasizing the importance of balancing the chemical equation first.
The first step is to write and balance the chemical equation for the combustion of propane: C3H8 + O2 -> CO2 + H2O. The carbon atoms are balanced first, then hydrogen, and finally oxygen, resulting in the balanced equation: C3H8 + 5O2 -> 3CO2 + 4H2O.
Given 2 moles of propane and 8 moles of oxygen, the limiting reactant is identified by dividing the moles of each reactant by its stoichiometric coefficient. For propane, 2 moles / 1 = 2. For oxygen, 8 moles / 5 = 1.6. Since oxygen has the lower ratio, O2 is the limiting reactant and propane is the excess reactant.
The theoretical yield of CO2 is calculated using both reactants. Starting with 2 moles of C3H8 and using the mole ratio (1 mole C3H8 : 3 moles CO2), 6 moles of CO2 can be formed if all propane reacts.
Using 8 moles of O2 and the mole ratio (5 moles O2 : 3 moles CO2), 4.8 moles of CO2 can be formed if all oxygen reacts. The smaller of the two yields (4.8 moles CO2) is the theoretical yield, confirming O2 as the limiting reactant.
If 4.5 moles of CO2 were actually collected, the percent yield is calculated as (actual yield / theoretical yield) * 100%. In this case, (4.5 moles / 4.8 moles) * 100% = 93.75%.
To find the amount of excess reactant (propane) left over, subtract the amount reacted from the initial amount. First, calculate how much propane reacted with the limiting reactant (8 moles of O2) using the mole ratio (5 moles O2 : 1 mole C3H8). This shows 1.6 moles of propane reacted. The initial 2 moles minus 1.6 moles reacted leaves 0.4 moles of propane left over.
A second example of benzene (C6H6) combustion with oxygen is introduced. The balanced equation for C6H6 + O2 -> CO2 + H2O is determined. To avoid fractions, all coefficients are doubled, resulting in 2C6H6 + 15O2 -> 12CO2 + 6H2O.
Given 50g of benzene, 160g of oxygen, and 30g of water collected, the 30g of water is identified as the actual yield. Molar masses for benzene (78 g/mol), oxygen (32 g/mol), and water (18 g/mol) are calculated for later use in stoichiometric calculations.
Starting with 50g of benzene, it's converted to moles, then to moles of water using the mole ratio (2 moles C6H6 : 6 moles H2O), and finally to grams of water. This calculation yields 34.6 grams of water.
Starting with 160g of oxygen, it's converted to moles, then to moles of water using the mole ratio (15 moles O2 : 6 moles H2O), and finally to grams of water. This calculation yields 36 grams of water. Since 34.6g is the smaller yield, it is the theoretical yield, and benzene is the limiting reactant.
With an actual yield of 30g of water and a theoretical yield of 34.6g of water, the percent yield is (30g / 34.6g) * 100% = 86.7%.
To find the amount of excess reactant (oxygen) left over, first calculate how much oxygen reacted with the limiting reactant (50g of benzene). Convert 50g of benzene to moles, then use the mole ratio (2 moles C6H6 : 15 moles O2) to find moles of O2 reacted, and finally convert back to grams of O2. This shows 153.8 grams of oxygen reacted. Subtracting this from the initial 160g of oxygen leaves 6.2 grams of oxygen left over.