1. Strategy for determining excess:
QUESTION:
Consider the reaction: N2 + 3 H2  2 NH3 For a mixture of 2.00 mol N2 and 3.00 mol H2, How much of the excess reactant would remain if the limiting reactant is consumed?
A mol, B mol, C mol, D mol
Strategy for determining excess:
assume one reactant is the limiting reactant and that it is completely consumed
calculate how much of the other reactant is also consumed; subtract from original amount to get remaining amount
If remaining amount is negative, you made the wrong assumption (redo). Otherwise, problem’s solved.
Consider the reaction: N2 + 3 H2  2 NH3 For a mixture of 2.00 mol N2 and 3.00 mol H2, How much of the excess reactant would remain if the limiting reactant is consumed?
A mol, B mol, C mol, D mol
PAUSE
CLICK
Here’s a strategy for solving this type of problem.
We assume one reactant is the limiting reactant and that it is completely consumed.
CLICK We then calculate how much of the other reactant would be consumed as well.
Then subtract that amount from the original amount of reactant.
If we get a negative remaining amount, then we know we made a wrong assumption.
So, we just go back to step 1 and assume that the other reactant is the limiting reactant.
Otherwise, we’re done.
CLICK CONTINUED ON NEXT SLIDE

2. Excess = 3.00 mol H2 - 6.00 mol H2 = -3.00 mol H2
QUESTION:
Consider the reaction: N2 + 3 H2  2 NH3 For a mixture of 2.00 mol N2 and 3.00 mol H2, How much of the excess reactant would remain if the limiting reactant is consumed?
A mol, B mol, C mol, D mol
3 mol H2
2.00 mol N2 x
= 6.00 mol H2
1 mol N2
Excess = 3.00 mol H mol H2 = mol H2
Let’ assume that nitrogen is the limiting reactant. Let’s calculate how much H2 would be consumed if
the nitrogen is completely consumed.
We start with the amount of nitrogen, 2.00 moles
CLICK
then multiply it by a conversion factor based on the coefficients in the balanced equation.
CLICK CLICK
The coefficient of hydrogen is 3
HIGHLIGHT coefficient in balanced equation
and the coefficient of nitrogen is 1
CALLOUT “Implied coefficient = 1” pointing to N2 in balanced equation
We find that the amount of hydrogen that would also be consumed is 6.00 moles
SHOW CANCELLATION OF UNITS
The remaining amount of hydrogen would be
the starting amount moles
HIGHLIGHT 3.00 mol H2 in problem and 3.00 mol H2 in calculation
minus
6.00 moles
HIGHLIGHT 6.00 mol H2 (both places)
which equals negative 3.00 moles of H2.
A negative amount is impossible. So our assumption that nitrogen is the limiting reactant is wrong.
Therefore, hydrogen must be the limiting reactant.
Let’s calculate the amount of nitrogen that would be used up assuming the hydrogen is completely consumed.
We start with 3.00 moles of hydrogen
CLICK times
a conversion factor based on the coefficients in the balanced equation
The coefficient of nitrogen is 1
CLICK CALLOUT “implied 1” pointing to left of N2 in balanced equation
The coefficient of hydrogen is 3.
CLICK HIGHLIGHT coefficient of H2
We find that the amount of nitrogen that would be used up assuming 3.00 moles of hydrogen are used up is 1.00 mole.
SHOW CANCELLATION OF UNITS CLICK
Therefore, the remaining amount of excess reactant is...
CLICK the starting amount moles
HIGHLIGHT 2.0 mol N2 in question and in calculation
minus 1.00 mole
HIGHLIGHT or 1.00 mole of hydrogen
The correct answer is B mole of hydrogen will remain unreacted if 2.00 moles of nitrogen are completely consumed.
The limiting reactant is hydrogen. The excess reactant is hydrogen.
CLICK PAUSE END RECORDING
1 mol N2
3.00 mol H2 x
= 1.00 mol N2
3 mol H2
Excess = 2.00 mol N mol N2 = 1.00 mol N2

3. Video ID: © 2008, Project VALUE (Video Assessment Library for Undergraduate Education), Department of Physical Sciences Nicholls State University Author: Glenn V. Lo Narrator: Funded by Louisiana Board of Regents Contract No. LA-DL-SELECT-13-07/08