Modern Chemistry, Chapter 9, Stoichiometry, pp 299-319. Holt, Winston & Rinehart, © 2009
Steps in solving stoichiometry problem.
1] Ensure chemical equation is balanced.
2] Write what is given over the substance(s), i.e., moles, grams, # atoms or molecules, volume
3] Write “?” *& what is desired over the substance(s) of interest.
4] If moles are not the only quantities provided, convert what is given into moles.
a) grams/ molar mass = moles
molar mass = sum of molar masses from periodic table
Do NOT consider the coefficient in determining the molar mass of substances.
b) # atoms or molecules/ Avogadro’s # = moles
Avogadro’s # = 6.022 x 1023 particles/ mole
c) at STP: Liter/ 22.4 L/mol = mole of gas
STP = standard temperature & pressure
Standard molar volume of all gases = 22.4 L/ mol
d) (volume)(density) = mass
(mL)(grams/ cm3) = grams
1 gram/ cm3 = 1 kg/ Liter
1000 g = 1 kg; 1000 mL = 1 Liter; cm3 = mL for water
Convert grams into moles as shown in #4a.
5] Determine the limiting reactant.
a) Make mole ratio from the coefficients in the balanced equation if moles are known for both.
Numerator = either reactant, but I prefer the expected excess reactant here.
Denominator = the other reactant, i.e., the expected limiting reactant
b) Multiply the mole ratio by the actual # of moles of reactant listed in the denominator.
Product denotes # moles required of the 1st reactant if the other is exhausted.
c) One can subtract the # moles needed from what is given to determine # moles excess.
6] Determine the # of moles of the unknown/ desired substance.
a) Set up another mole ratio from the coefficients of the balanced equation
Numerator = the moles of the “?” substance
Denominator = moles of the limiting reactant
b) Multiply this mole ratio by the actual # moles of limiting reactant that is given or calculated.
c) Product = # moles of substance in question
7] Convert moles into desired quantity.
a) moles x grams/mole = grams
b) moles x 6.022 x 1023 particles/ mole = # particles
c) grams/ grams/cm3 = cm3 = mL = 0.001 L
d) moles x 22.4 L/ mol = Liter of gas at STP