Limiting and Excess Reactants
Frequently Asked Questions (FAQ’s) about Limiting Reactant:
1. What is it? It’s the control of the reaction,  It gives the right answers.
2. Why does it work? It’s the lowest amount, it runs out first thus stopping the reaction.
3. When do we know to find it? If more than 1 reactant is given at the start
4. How do we find it? (1) find moles (2) divide by coefficient (# in front) (3) lowest
5. What do you do with it once you find it? Use it to start every problem!
2 HMnO4(aq) + 5 Na3AsO3(s) + 19 HNO3(l)  2 Mn(NO3)2(s) + 5 H3AsO4(l) + 15 NaNO3(aq) + 3 H2O(g)
(120)
(192)
(63)
(179)
(142)
(85)
(18)
Density=1.35g/mL
Density=1.7g/mL
molar masses
(g/mol)
1. If you start with 45 g of HMnO4 and 32 g of Na3AsO3, how many atoms of NaNO3 will be
produced? Don’t forget to find the limiting reactant and use it to find atoms of NaNO3.
2. If you mix 450 g HMnO4 with 4 X 1025 atoms Na3AsO3 and 1.5 liters of HNO3. Find the amount
of NaNO3, in grams, that will form.
Leftover or Excess Reactants—does it make sense, that if the limiting reactant runs out first;
then, none will be left over when the reaction stops? But there would be some of the non-limiting
reactants left over. To find leftovers, you start with the limiting reactant and calculate how much of
the other reactant(s) would be USED. Then subtract the used amount from the given amount to
find leftover.
2 HMnO4(aq) + 5 Na3AsO3(s) + 19 HNO3(l)  2 Mn(NO3)2(s) + 5 H3AsO4(l) + 15 NaNO3(aq) + 3 H2O(g)
(120)
(192)
(63)
(179)
(142)
(85)
(18)
Density=1.35g/mL
Density=1.7g/mL
molar masses
(g/mol)
3. Find the amount, in grams, of leftover reactants from # 1.
4. Find the amount of leftover reactants from #2.