stoichiometry. stoichiometry comes from the greek words for “element” and “measure”....

Stoichiometry

Stoichiometry comes from the Greek words for “element” and “measure”.

Stoichiometry takes information for one element/compound in a reaction and allows for the calculation of the amount of a different element/compound in the same reaction.

Obviously the place to start is…

1. Predict and write a balanced chemical equation.

1. Predict and write a balanced chemical equation.

2. Start a RR track and convert the given information to moles (if not already as moles).

1. Predict and write a balanced chemical equation.

2. Start a RR track and convert the given information to moles (if not already as moles).

a. Use the molar mass (from the PT) to change grams → moles

1. Predict and write a balanced chemical equation.

2. Start a RR track and convert the given information to moles (if not already as moles).

a. Use the molar mass (from the PT) to change grams → moles

b. Use 6.022 x 1023 to change atoms/molecules/formula units → moles

1. Predict and write a balanced chemical equation.

2. Start a RR track and convert the given information to moles (if not already as moles).

a. Use the molar mass (from the PT) to change grams → moles

b. Use 6.022 x 1023 to change atoms/molecules/formula units → moles

c. Use 22.4 L (for gases only) to change L → moles

1. Predict and write a balanced chemical equation.

2. Start a RR track and convert the given information to moles (if not already as moles).

a. Use the molar mass (from the PT) to change grams → moles

b. Use 6.022 x 1023 to change atoms/molecules/formula units → moles

c. Use 22.4 L (for gases only) to change L → moles

d. For water and dilute solutions, 1 mL = 1 g

3. Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation

a. Mole # of unknownMole # of known

3. Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation

a. Mole # of unknownMole # of known

4. Convert the moles of the unknown to the desired unit (see possible conversion factors in step 2)

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1:

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid1.00 x 102 g K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid1.00 x 102 g K 1 mol K

39.1 g K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4

=39.1 g K 6 mol K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4

= 0.853 mol H3PO439.1 g K 6 mol K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

b. Find the mass of phosphoric acid

1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4

39.1 g K 6 mol K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

b. Find the mass of phosphoric acid

1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4

39.1 g K 6 mol K

1.00 x 102 g K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

b. Find the mass of phosphoric acid

1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4

39.1 g K 6 mol K

1.00 x 102 g K 1 mol K

39.1 g K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

b. Find the mass of phosphoric acid

1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4

39.1 g K 6 mol K

1.00 x 102 g K 1 mol K 2 mol H3PO4

39.1 g K 6 mol K

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

b. Find the mass of phosphoric acid

1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4

39.1 g K 6 mol K

1.00 x 102 g K 1 mol K 2 mol H3PO4 98.0 g H3PO4=

39.1 g K 6 mol K 1 mol H3PO4

Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning

Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2

a. Find the moles of phosphoric acid

b. Find the mass of phosphoric acid

1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4

39.1 g K 6 mol K

1.00 x 102 g K 1 mol K 2 mol H3PO4 98.0 g H3PO4= 83.5 g H3PO4

39.1 g K 6 mol K 1 mol H3PO4

c. Find the formula units of phosphoric acid

c. Find the formula units of phosphoric acid1.00 x 102 g K

c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K

39.1 g K

c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4

=39.1 g K 6 mol K 1 mol H3PO4

c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4

= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K

39.1 g K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4=

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

e. Find the mass of potassium phosphate

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

e. Find the mass of potassium phosphate1.00 x 102 g K

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K

39.1 g K

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K 2 mol K3PO4

39.1 g K 6 mol K

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K 2 mol K3PO4 212 g K3PO4

=39.1 g K 6 mol K 1 mol K3PO4

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

c. Find the formula units of phosphoric acid

d. Find the moles of potassium phosphate

e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K 2 mol K3PO4 212 g K3PO4

= 181 g K3PO439.1 g K 6 mol K 1 mol K3PO4

1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4

1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4

39.1 g K 6 mol K

f. Find the mass of the hydrogen

f. Find the mass of the hydrogen1.00 x 102 g K

f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K

39.1 g K

f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K 3 mol H2

39.1 g K 6 mol K

f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2

=39.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2

= 2.58 g H239.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K

39.1 g K

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2

39.1 g K 6 mol K

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2=

39.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

h. Find the molecules of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

h. Find the molecules of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

h. Find the molecules of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K

39.1 g K

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

h. Find the molecules of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2

39.1 g K 6 mol K

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

h. Find the molecules of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 6.022 x 1023 molecules H2=

39.1 g K 6 mol K 1 mol H2

f. Find the mass of the hydrogen

g. Find the liters of hydrogen

h. Find the molecules of hydrogen

1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2

39.1 g K 6 mol K 1 mol H2

1.00 x 102 g K 1 mol K 3 mol H2 6.022 x 1023 molecules H2= 7.70 x 1023 molecules H239.1 g K 6 mol K 1 mol H2

Limiting Reactant

Rarely in real life does a reaction start with exactly proportional amounts of the reactants, as the smallest differences in mass still add up to more than trillions of atoms, and often there are big differences in the amounts of the reactants available.

To determine which reactant will limit the reaction, the initial amount of all reactants must be known. Then take each react in turn and solve for the same product (it will not matter which product, so choose something convenient). The reactant the can produce the smallest amount of products is the limiting reactant, and all the others are the excess reactants.

Example: 10.0 grams of sodium is added to 50.0 grams of water. Which reactant is the limiting reactant and which one is the excess reactant?

Example: 10.0 grams of sodium is added to 50.0 grams of water. Which reactant is the limiting reactant and which one is the excess reactant?

Step 1: 2Na + 2HOH → 2NaOH + H2

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na

23.0 g Na

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH

23.0 g Na 2 mol Na

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH=

23.0 g Na 2 mol Na 1 mol NaOH

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

50.0 g H2O

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

50.0 g H2O 1 mol H2O

18.0 g H2O

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

50.0 g H2O 1 mol H2O 2 mol NaOH

18.0 g H2O 2 mol H2O

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

50.0 g H2O 1 mol H2O 2 mol NaOH 40.0 g NaOH=

18.0 g H2O 2 mol H2O 1 mol NaOH

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

50.0 g H2O 1 mol H2O 2 mol NaOH 40.0 g NaOH= 111 g NaOH

18.0 g H2O 2 mol H2O 1 mol NaOH

Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

Na is limiting and thus H2O is excess.

10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH

23.0 g Na 2 mol Na 1 mol NaOH

50.0 g H2O 1 mol H2O 2 mol NaOH 40.0 g NaOH= 111 g NaOH

18.0 g H2O 2 mol H2O 1 mol NaOH

How much excess is there?Step 1: Starting with the limiting...

How much excess is there?Step 1: Starting with the limiting...

10.0 g Na

How much excess is there?Step 1: Starting with the limiting...

10.0 g Na 1 mol Na

23.0 g Na

How much excess is there?Step 1: Starting with the limiting...

10.0 g Na 1 mol Na 2 mol H2O

23.0 g Na 2 mol Na

How much excess is there?Step 1: Starting with the limiting...

10.0 g Na 1 mol Na 2 mol H2O 18.0 g H2O=

23.0 g Na 2 mol Na 1 mol H2O

How much excess is there?Step 1: Starting with the limiting...

10.0 g Na 1 mol Na 2 mol H2O 18.0 g H2O= 7.83 g H2O used

23.0 g Na 2 mol Na 1 mol H2O

How much excess is there?Step 1: Starting with the limiting...

Step 2: 50.0 g H2O – 7.83 g H2O = 42.17 g H2O

10.0 g Na 1 mol Na 2 mol H2O 18.0 g H2O= 7.83 g H2O used

23.0 g Na 2 mol Na 1 mol H2O

How many grams of base are produce?

How many grams of base are produce?17.4 g NaOH

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?10.0 g Na

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?10.0 g Na 1 mol Na

23.0 g Na

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2

23.0 g Na 2 mol Na

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2 22. 4 L

23.0 g Na 2 mol Na 1 mol H2 gas

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2 22. 4 L 1000 mL

= 23.0 g Na 2 mol Na 1 mol H2 gas 1 L

How many grams of base are produce?17.4 g NaOH

How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2 22. 4 L 1000 mL

= 4870 mL23.0 g Na 2 mol Na 1 mol H2 gas 1 L