Thinking of aqueous solutions, the solute is the material (salt, sugar, etc.) placed into the solvent (water). Usually there is very little solute and a lot of solvent. The concentration of the solute in the solvent is an expression of just how much solute is placed into the solvent. The solute and solvent together make the solution.
Molarity expresses the amount of solute per liter of solution
and has units of moles of solute per liter of solution (moles
solute/L solution). The molarity is a conversion factor between
moles of solute and liters of solution. Knowing the volume
(liters) of solution and the molarity is enough to determine
the moles of solute. If the solute is a reactant, these moles
can be used in limiting reacatant problems to determine the
amount of product expected from the reaction.
Molarity as a Conversion Factor
How many moles of HCl are in 0.5 L of a 1.2 M HCl solution?
| ( |
0.5 L HCl Soln
| ) | ( |
|
1.2 mole HCl
| | | |
1 L HCl Soln
|
| ) | |
= 0.6 mole HCl
|
How many liters of a 1.2 M HCl solution are needed to provide 0.24 mole HCl?
| ( |
0.24 mole HCl
| ) | ( |
|
1 L HCl Soln
| | | |
1.2 mole HCl
|
| ) | |
= 0.2 L HCl Soln
|
Using Molarity to do Limiting Reactant Problems
Consider the following representation of the chemical reaction between NaCl and AgNO3:
NaCl(aq) + AgNO3(aq)
→
NaNO3(aq) + AgCl(s)
How many grams of AgCl(s) will be produced if 0.2 L of 1.3 M NaCl solution is added to 0.3 L of 0.6 M AgNO3?
This is a limiting reactant problem. First determine the moles of reactants initially present (using the molarity conversion factor). Then determine the limiting reactant (using mole ratios from the balanced equation). Finally, determine the moles of product formed (from balanced equation) and grams of product formed (from periodic table gram-mole conversion factor).
NaCl initially available for reaction:
| ( |
0.2 L NaCl Soln
|
) |
( |
|
1.3 mole NaCl
|
| |
|
1 L NaCl Soln
|
|
) |
|
= 0.26 mole NaCl
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AgNO3 initially available for reaction:
| ( |
0.3 L AgNO3 Soln
|
) |
( |
|
0.6 mole AgNO3
|
| |
|
1 L AgNO3 Soln
|
|
) |
|
= 0.18 mole AgNO3
|
AgNO3 needed to react with all of the NaCl:
| ( |
0.26 mole NaCl
|
) |
( |
|
) |
|
= 0.26 mole AgNO3 needed
|
There are actually 0.18 moles of AgNO3 available, which means
that there is not enough AgNO3 to react with all of the NaCl. This means that the AgNO3 will
all be used up and AgNO3 is the limiting reactant.
Once the limiting reactant is determined, the moles of product can be determined. Use the mole ratios from the balanced chemical equation to convert from moles of limiting reactant to moles of product and then use the gram-mole conversion factor from the periodic table to determine the grams of product formed.
| ( |
0.18 mole AgNO3
| ) | ( |
| ) | ( |
|
143.32 g AgCl
| | | |
1 mole AgCl
|
| ) | |
= 25.8 g AgCl
|
Another Limiting Reactant Example
Question: How many grams of BaSO4 and how many grams of KNO3 will be produced when 400 mL of 1.8 M BaNO3 solution reacts with 300 mL of a 1.2 M K2SO4 solution?
Write the balanced equation:
Ba(NO3)2(aq) + K2SO4(aq)
→
BaSO4(s) + 2KNO3(aq)
The only extra step in this problem is to convert from milliliters to liters (which I have done in my head).
BaNO3 initially available for reaction:
| ( |
0.4 L BaNO3 Soln
| ) | ( |
|
1.8 mole BaNO3
| | | |
1 L BaNO3 Soln
|
| ) | |
= 0.72 mole BaNO3
|
K2SO4 initially available for reaction:
| ( |
0.3 L K2SO4 Soln
| ) | ( |
|
1.2 mole K2SO4
| | | |
1 L K2SO4 Soln
|
| ) | |
= 0.36 mole K2SO4
|
K2SO4 needed to react with all of the BaNO3:
| ( |
0.72 mole BaNO3
| ) | ( |
|
1 mole K2SO4
| | | |
1 mole BaNO3
|
| ) | |
= 0.72 mole K2SO4 needed
|
There are actually 0.36 moles of K2SO4 available, which means
that there is not enough K2SO4 to react with all of the NaCl. This means that the K2SO4 will
all be used up and K2SO4 is the limiting reactant. Use the initial amount of K2SO4 to determine the amount of products produced:
| ( |
0.36 mole K2SO4
| ) | ( |
|
1 mole BaSO4
| | | |
1 mole K2SO4
|
| ) | ( |
|
217.93 g BaSO4
| | | |
1 mole BaSO4
|
| ) | |
= 218.29 g BaSO4
|
| ( |
0.36 mole K2SO4
| ) | ( |
| ) | ( |
| ) | |
= 72.8 g KNO3
|
There will be maximum of 218.29 g BaSO4 and 72.8 g KNO3 produced.