1. 25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M
H2SO4. Determine the moles of the excess reactant remaining and
its new concentration.
2NaOH + H2SO4  2H2O Na2SO4
25.0 mL
14.7 mL
Just pick a
product

2. 25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M
H2SO4. Determine the moles of the excess reactant remaining and
its new concentration.
2NaOH + H2SO4  2H2O Na2SO4
25.0 mL
14.7 mL
Let’s
choose
H2O

3. 25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M
H2SO4. Determine the moles of the excess reactant remaining and
its new concentration.
2NaOH + H2SO4  2H2O Na2SO4
25.0 mL
14.7 mL
? moles
NaOH
25.0 mL
Based on:
NaOH
= mol H2O
0.0125
Based on:
H2SO4
= mol H2O
0.0103
Theoretical yield?
Limiting Reactant?

4. 25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M
H2SO4. Determine the moles of the excess reactant remaining and
its new concentration.
2NaOH + H2SO4  2H2O Na2SO4
25.0 mL
14.7 mL
? moles
NaOH
25.0 mL
Based on:
NaOH
= mol H2O
0.0125
Based on:
H2SO4
= mol H2O
0.0103
– = moles of H2O that can be formed
from the excess NaOH.

5. SO: 0.0022 mol H2O = mole NaOH 0.0022 NOW: 0.0022 mol NaOH 0.055
25.0 mL of 0.50 M NaOH were combined with 14.7 mL of 0.35 M H2SO4. Determine the moles of the excess reactant remaining and its new concentration.
2NaOH + H2SO4  2H2O Na2SO4
25.0 mL
14.7 mL
? moles
– = moles of H2O that can be formed in excess
from the excess NaOH.
SO:
mol H2O
= mole NaOH
0.0022
NOW:
mol NaOH
0.055
39.7 x 10-3 Lsolution