1. 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 = 100.g / 55.99g/mol
MOLES = mol KOH
MOLES = MASS/GFM H2SO4
MOLES = 100. g / 98.00g/mol
MOLES = mol
H2SO4 = =
KOH
.50 <
The actual ratio is larger than the theoretical, therefore the actual ratio fraction is too large, the denominator is to small, KOH limiting.

2. 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
mol
1.0204mol
0.0 C
-2X -X +X
+2X E –X

3. DETERMINING THE VALUE OF “X” X IS DETERMINED BY THE LIMITING REAGENT.
IN THIS PROBLEM KOH, mol = 2X FROM THE LIMITING REAGENT KOH. X = MOL.
Quantities of all species are calculated from x:
THE EXCESS REAGENT IS SULFURIC ACID, WHICH IS –X
– 0.893
MOLE OF H2SO4 REMAINS
2. K2SO4 PRODUCED IS X, MOLE
3 WATER IS 2X, 2(0.893) = MOLE.
MOLES = MASS/GFM K2SO4
0.893mol= MASS/ g
MASS = g K2SO4
154. g
MOLES H2O= MASS/GFM H2O
1.786 MOLE = MASS/ : MASS = g = 32.1g