2 KOH (s) + H2SO4 (g) K2SO4 (aq) + 2 H2O (g)
In a particular experiment, the reaction of 100. g KOH with 100.g H2SO4
•
1) Calculate the MASS of all species present at the end of the reaction.
FIRST FIND YOUR LIMITING REAGENT
MOLES = MASS/GFM KOH
MOLES = MASS/GFM H2SO4
MOLES = 100.g / 55.99g/mol
MOLES = 100. g / 98.00g/mol
H2SO4 = 1 = 1.0204
KOH
2
.50 <
MOLES = 1.786033 mol KOH
MOLES = 1.0204mol
1.786033
The actual ratio is larger than the
theoretical, therefore the actual ratio
fraction is too large, the denominator is
.57
to small,
KOH limiting.
ICE CHART SOLUTION SKILL
THE ICE CHART WILL ALLOW YOU TO CALCULATE MOLES OF ALL
SPECIES AFTER THE REACTION IS COMPLETED.
MOLES OF ALL PRODUCTS.
MOLES OF ALL EXCESS REACTANTS REMAINNING.
THE FASTEST WAY TO APPROACH A REACTION IN WHICH YOU
ARE CALCULATING MANY SUBSTANCES AT ONCE.
2KOH(s) +
H2SO4(g) K2SO4(aq) +2H2O(g)
I 1.786033 mol
C -2X
1.0204mol
E
1.0204 –X
0.0
-X
0.0
0.0
+X
+2X
+X
+2X
DETERMINING THE VALUE OF “X”
1. X IS DETERMINED BY THE LIMITING REAGENT.
2. IN THIS PROBLEM KOH, 1.786033 mol = 2X
FROM THE LIMITING REAGENT KOH. X = 0.893
MOL.
Quantities of all species are calculated from x:
1. THE EXCESS REAGENT IS SULFURIC ACID, WHICH
IS 1.0204 –X
MOLES = MASS/GFM K2SO4
1.0204 – 0.893
0.893mol= MASS/173.99 g
0.1274 MOLE OF H2SO4 REMAINS MASS = 153.87 g K2SO4
154. g
2. K2SO4 PRODUCED IS X, 0.893 MOLE
3 WATER IS 2X, 2(0.893) = 1.786 MOLE.
MOLES H2O= MASS/GFM H2O
1.786 MOLE = MASS/17.99 : MASS = 32.103g = 32.1g