1. Solutions and Stoichiometry

2. We can apply our knowledge of stoichiometry to solutions
We can apply our knowledge of stoichiometry to solutions. Concepts like mole ratios and limiting reagents are just as useful when dealing with solutions.

3. Write a balanced chemical equation.
Strategy:
Write a balanced chemical equation.
Determine the number of moles of the reactants. You will need to use: c=n/V and/or n=m/mm
Use c=n/V if aqueous (solution), Use n=m/mm if solid, liquid or gas (pure)
Determine the limiting reagent.
Use the limiting number of moles in a ratio to find unknowns.
Convert back from moles to mass

4. Step 1: Balance the chemical equation.
Example 1: Calculate the volume of 1.12 mol/L barium iodide you would need to react completely with 150 mL of 1.6 mol/L silver nitrate.
Step 1: Balance the chemical equation.
BaI2(aq) + 2AgNO3(aq)  Ba(NO3)2(aq) + 2AgI(s)

5. Determine the number of moles of the reactants.
We are given the C and V values for silver nitrate, which we can use to find “n”. Silver nitrate is a solution and therefore not pure. You cannot use molar mass to find n.
n=cV
=(1.6 mol/L)(0.150L)
=0.24 mol of AgNO3

6. Use the mole ratio to determine how many moles of BaI2 would be required.=
2x = 0.24
x= 0.12 moles of BaI2
1 BaI2
2 AgNO3 x
0.24

7. V = n/C = 0.12 moles/1.12 mol/L = 0.11L = 110 mL
We can now find the V of BaI2 required by using n and C. Notice barium iodide is also a solution (aqueous) so c=n/V is the best option.
V = n/C
= 0.12 moles/1.12 mol/L
= 0.11L
= 110 mL

8. Mg(s) + 2HCl(aq)  H2(g) + MgCl2(aq)
Example 2: Putting magnesium into hydrochloric acid produces hydrogen gas and magnesium chloride. If you put a 0.56g strip of magnesium into 100 mL of 1.00 mol/L HCl, determine the concentration of magnesium chloride that will be present after the reaction. (assume the volume remains constant). Notice Mg is a solid so n=m/mm is OK for this pure substance but not for HCl (aq)
Balance the equation.
Mg(s) HCl(aq)  H2(g) + MgCl2(aq)

9. We are given information to determine the number of moles of both reactants:
nMg = m/mm
= 0.56/24.31
= mol Mg
nHCl = cV
= 1.00 mol/L(0.100 L)
= mol HCl

10. Determine the limiting reagent:
Case 1 =
x = mol MgCl2
Case 2 =
x = mol MgCl2
0.023 x
0.100 x Mg
MgCl2
2 HCl
MgCl2
Mg is the limiting reagent since it predicts less product formed.

11. Using Case 1, we calculated 0.023 mol of MgCl2 produced.
Use the moles of Mg in the rest of the calculations, because it is the limiting reagent.
Using Case 1, we calculated mol of MgCl2 produced.
Using the volume of the solution of 100 mL, calculate the concentration. C=n/V is the best equation for the solution since it is a mixture.
C = n/V
= 0.023/0.100L
= 0.23 mol/L