SCH 3U1
7.4 LIMITING REAGENTS
EX 1 If 34.7 g of sodium carbonate was reacted with 41.2 g of calcium chloride, how much sodium chloride will be produced by this reaction?
STEP 1 - BALANCED CHEMICAL EQUATION
__ Na2CO3(aq) + __ CaCl2(aq) → __CaCO3(s) + __ NaCl(aq)
STEP 2 - FIND THE LIMITING REAGENT
mol of Na2CO3
n= m
MM
mol of CaCl2
n= m
MM
n=
n=
g
g/mol
n _____ mol
g
g/mol
n = _____ mol
SINCE THE MOLES OF SODIUM CARBONATE IS THE LOWER NUMBER, IT IS THE LIMITING REAGENT.
STEP 3 - FIND THE MASS OF SODIUM CHLORIDE
To find the moles of sodium chloride:
Na2CO3 =
mol
NaCl
x
x = _____ mol of NaCl
mass of NaCl = n x MM
mass of NaCl = _____ mol x _____ g/mol
mass of NaCl = _____ g.
Ex 2. If 21.0 g of aluminum hydroxide reacts with 40.7 g of sulphuric acid, what is the mass of water produced?
STEP 1 - BALANCED CHEMICAL EQUATION
__ Al(OH)3(aq) + __ H2SO4(aq) ---> __ Al2(SO4)3(aq) + __ H2O(l)
STEP 2 - FIND THE LIMITING REAGENT
mol of Al(OH)3
mol of H2SO4
n=
n= m
MM
m
MM
n = 21.0 g
n = 40.7 g___
g/mol
g/mol
n = _____ mol
n = _____ mol
In order to compare the relative number of moles:
n of Al(OH)3 =
/
n of H2SO4 =
n = _______
/
n = ______
Therefore by a direct comparison using the relative number of moles, the limiting reagent is
STEP 3 - FIND THE MASS OF WATER
If aluminum hydroxide has _______ mol then water can be calculated:
Al(OH)3 =
mol
H2O
x
mass of H2O = n x MM
mass of H2O = ____ mol x ______ g/mol
mass of H2O = _____ g
x = _____ mol of water
.