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Stolchlometry III. Calculations Based on Concentrations of Solutions
NaOH, we will know the number of moles of H 2 SO 4 used by the Na 2 CO 3 This will
tell us how much Na 2 CO 3 must have been present originally
Total moles H 2 SO 4 at outset = (o 06322
mol e
s H 2 SOA (Q Q5QQQ
= 0003161 moles H 2 SO 4
moles of H 2 SO 4 used by NaOH
= 0 ,079
, "" (0 00936 liters) .,,
,
V
liter
/
\2 moles
= 0 000505 moles H 2 SO 4
Moles of H 2 SO 4 not used by NaOH = 0 003,61 - 0 000505
= 0 002656 moles H 2 SO 4 used by Na 2 CO 3
The chemical equation shows that 1 mole of Na 2 CO 3 is used per mole of H 2 SO 4 , so
weight of Na-jCQs present
= (0 002656 moles HzSOJ ('
mole ^ff^
3 ) (,06 0 ^ff^' )
V 1 mole H 2 SO 4 / \
mole Na 2 COj/
= 0 2815 g Na 2 CO 3 in sample
„ „ „„
02815eNa,CO
0 33,2 g sample * >°° = ^ 00%
PROBLEM:
An iron ore is a mixture of Fe 2 O 3 and inert impurities One method of iron analysis
is to dissolve the ore sample in HCl, convert all of the iron to the Fe+ state with
metallic zinc, then titrate the solution with a standard KMnO 4 solution The reac
tion is
KMnO 4 + 5FeCl 2 + 8HC1 -> KCl + MnCl 2 + 5FeClj + 4H 2 O
MnO 4 + 5Fe
J+ + 8H
+ ^ Mn
2+ + 5Fe
3+ + 4H 2 O
At the endpomt, there is no more FeCl 2 for reaction, so the further addition of
purple KMnO 4 will make the whole solution purple because it can no longer be
converted to colorless MnCl 2 If aO 3778 g ore sample requires 38 60 ml of 0 02 1 0
1
^
M KMnO 4 for titration, calculate the percentage of iron in the original sample
SOLUTION:
1 From the given amount of KMnO 4 , calculate the moles of KMnO 4 used
mole MnO«
moles KMn0 4 used = (o 02105
mole ^MnO«j (Q
= 8 125 x 10
4 moles KMnO 4