Stoichiometry and Solutions
Name_________________________
Date____________ Block____
How many mL of 0.1 M NaOH(aq) are needed to neutralize 10.0 mL of 0.1 M H2SO4 (aq)?
1) Write the balanced chemical equation:
2 NaOH(aq)
+
H 2SO4(aq)
→
Na2SO4 (aq) + 2 HOH(l)
2) Write the given molarity and volume underneath the involved substances; identify the “to find”:
2 NaOH(aq)
? mL, 0.1 M
+
H2SO4(aq)
10.0 mL, 0.1 M
→
Na2SO4 (aq) + 2 HOH(l)
3) In stoichiometry, we used known moles of substances and the mole ratios to calculate unknown amounts
of other substances. Here, we will do the same types of calculations. The only difference is that there are
additional calculations: we need to calculate how many moles of a substance are present in the volume given.
For this calculation, we use the molarity of the solution.
First calculate how many moles of H2SO4 are in 10.0 mL of 0.1 M solution:
M=
# moles solute
1.0 L of solution
# moles solute = molarity x (# L of solution)
# moles H2SO4 = 0.1 moles H2SO4
1.0 L solution
# moles H2SO4
=
X
0.010 L solution
0.001 moles H2SO4 present in 10.0 mL 0.1 M H2SO4(aq)
Second, calculate how many moles of NaOH are needed to react with 0.001 moles H 2SO4 using the mole ratio:
0.001 moles H2SO4
X
2 moles NaOH
1 moles H2SO4
= 0.002 moles NaOH needed
Third, calculate how many mL of 0.1 M NaOH contain 0.002 moles NaOH using the molarity:
0.1 M NaOH = 0.002 moles NaOH
? L of solution
0.1 mole NaOH
1.0 L of solution
= 0.002 moles NaOH
? L of solution
(0.1 mole NaOH)(? L of solution) = (1.0 L of solution)(0.002 moles NaOH)
? L of solution = (1.0 L of solution)(0.002 moles NaOH)
0.1 mole NaOH
? L of solution = 0.02 L of 0.1 M NaOH needed to neutralize 10.0 mL of 0.1 M H 2SO4
to change L to mL: 0.02 L
X
1000 mL = 20 mL of 0.1 M NaCl needed to neutralize 10.0 mL of 0.1 M H2SO4
1L
SHOW ALL work for items requiring calculations!!!!!!
DO all work on separate paper.
Use the following UNBALANCED equation for the questions below:
Pb(NO3)2 (aq) +
KI (aq) →
KNO3 (aq)
+
PbI2 (s)
1. Correctly balance the equation.
2. How many mL of 0.1 M KI are required to react completely with 25.0 mL 0.5 M
lead(II) nitrate?
3. How many grams of lead(II) iodide will be formed if 25.0 mL of 0.01 M potassium iodide
react completely?
4. What volume of 1.5 M lead(II) nitrate will react completely with 5.0 mL 2.0 M
potassium iodide?
5. What volume of 1.0 M potassium iodide is required to react completely with 0.001 M
lead(II) nitrate?
6. What mass of lead(II) iodide will form if 0.5 mL 0.25 M potassium iodide reacts
completely?
7. What volume of 1.20 M potassium iodide will react completely with 10.0 mL 0.001 M
lead(II) nitrate?
8. What volume of 0.002 M lead(II) nitrate contains 0.15 g lead(II) nitrate?
9. What volume of 1.5 M potassium nitrate contains 1.5 g potassium nitrate?
10. What volume of 0.5 M potassium iodide contains 3.4 g potassium iodide?