STOICHIOMETRY

- writing balanced chemical equations

- predicting products for chemical reactions

- converting mass to # of moles

- converting # of moles to mass


1.Mole to Mole Stoichiometry

Every problem in stoichiometry is based on a chemical equation and its interpretation of moles.
The equation :
2 C + O₂ -------> 2 CO
- may be interpreted as saying that 2 moles of carbon combine with 1 mole of oxygen to form 2 moles of carbon monoxide. The coefficients in the equation express the relative reacting quantities in terms of moles.
From this it follows that if the reaction quantitiy of carbon is 4 moles (twice as much as is indicated by the equation) the reaction quantities of oxygen is 2 moles and 4 moles of carbon monoxide:
4 C + 2 O₂ -------> 4 CO
Likewise if the reacting quantity of carbon is half as much as indicated by the original equation, what would the number of moles be for O2 and CO? .5 mol O2 1 mol CO

Example 1:

Balance the following equation and fill in the table:

N2	+	3 H2	-->	2 NH3
6 mol		18 mol		12 mol
2 mol		6 mol		4 mol
3 mol		9 mol		6 mol

Example 2:

Methane burns in oxygen to produce carbon dioxide and water. If 8.2 moles of methane are burned in the reaction, how many moles of water will be produced?

Solution

Step 1

 Write the balanced equation for the reaction and the mole ratio for the given and required substances.

                          Step 2

Write the given and the unknown beneath the appropriate substance.

 

CH4	+	2O2	→	CO2	+	2H2O
1 mol						2 mol
8.2 mol						X mol

 

Step 3

 Set up a ratio:

 

1 mol	=	2 mol
8.2 mol		X mol

 Step 4

 Solve for "X". Use cross multiplication.
 1 mol x X mol = 8.2 mol x 2 mol ; X = 16.4 mol 
Therefore the number of moles of H2O formed from 8.2 mol of methane is 16.4 mol
Example 3:
Propane burns in oxygen to produce carbon dioxide and water. If 0.2 moles of propane are burned in the reaction, how many moles of oxygen is required?

Solution

Step 1

 Write the balanced equation for the reaction and the mole ratio of the given and required substances.

 

C3H3	+	5 O2	→	3 CO2	+	4 H2O
1 mol		5 mol

Step 2

 Write the given and the unknown beneath the appropriate substance.

 

C3H3	+	5 O2	→	3 CO2	+	4 H2O
1 mol		5 mol
0.2 mol		X mol

Step 3 X

 Set up a ratio:

 

1 mol	=	5 mol
0.2 mol		X mol

 Step 4

 Solve for "X". Use cross multiplication.
 1 mol x X mol = 0.2 mol x 5 mol  ; X = 1 mol  
Therefore, 1 mole of oxygen is required when 0.2 mol of propane are burned

Step 1

 Write the equation, the moles and masses of the substances beneath the required substances.

 

C3H8	+	5O2	→	3 CO2	+	4 H2O
1 mole		5 moles
44 g		5 (32) g

 Step 2

Write the given value (12.9 g propane) and "X" for the unknown (oxygen) beneath the appropriate substance.
 

C3H8	+	5O2	→	3 CO2	+	4 H2O
1 mole		5 moles
44 g		5 (32) g
12.9 g		X

 Step 3

Set up a ratio.

44 g	=	5 (32)g
12.9 g		X

 Step 4

Solve for "X". Use cross multiplication.
 44 g x X g = 12.9 g x 160 g
 X = (2064 g ) / 44 g = 46.9 g
Therefore 12.9 grams of propane will produce 46.9 grams of oxygen.

3.Mass to Molecule

A mass to molecule probelm is one in which the mass of a reactant or product is given. You are then asked to calculate the the number of molecules of another reactant required or the mass of another product formed.

Example 1:
Iron is prepared by reacting iron ore (Fe₂O₃) with carbon monoxide (CO). Iron and carbon dioxide result from the reaction. How many molecules of CO would react with 600 g of iron ore?

Step 1

Write the balanced equation, mole ratio, and masses beneath the required substances.

 

Fe2O3	+	3 CO	→	2 Fe	+	3 CO2
1 mole		3 moles
159.6 g		3(6.02x1023)molecules

Step 2

Write the given and the unknown beneath the appropriate substance.

Fe2O3	+	3 CO	→	2 Fe	+	3 CO2
1 mole		3 moles
159.6 g		3(6.02x1023)molecules
600g		X molecules

Step 3

Set up a ratio:

159.6 g	=	3(6.02x1023)molecules
600 g		X molecules

Step 4

Solve for "X". Use cross multiplication.
159.6 g x X g = 600 g x 3(6.02x10²³)molecules
X =6.79 x 10²³ molecules
Therefore 6.79 x 10²³ molecules of carbon monoxide will react with 600g of iron ore.

4.Volume to Volume Stoichiometry

A volume to volume probelm is one in which the volume of a reactant or product is given. You are then asked to calculate the volume of another reactant required or the mass of another product formed.
Example 1:

Nitrogen and hydrogen react to produce ammonia. What volume of ammonia could be produced from 300 liters of nitrogen at 200°C and 150 kPa? Assume that enough hydrogen is available for a complete reaction.

Sample solution:

step 1

Write a balanced chemical equation and the number of moles and the volumes of all the substances in the equation.

 

N2	+	3 H2	→	2 NH3
1 mol		3 mol		2 mol
22.4 L (STP)				2 (22.4) L (STP)

Step 2

Write the given value and "X" for the unknown beneath the appropriate substance. 

 

N2	+	3 H2	→	2 NH3
1 mol		3 mol		2 mol
22.4 L (STP)				2 (22.4) L (STP)
300L at 473K and 150 kPa				X L at 473K and 150 kPa

Step 3

Set up a ratio.

22.4 L (STP)	=	2 (22.4) L (STP)
300 L(at 473K and 150 kPa)		X L  (at 473Kand    150 kPa)

Step 4

Solve for "X". Use cross multiplication.
22.4 L(STP) x X L(473K, 150kPa) = 300 L x 2 (22.4) L (STP)
X = 600 L
Therefore 600 L of ammonia gas will form at 200°C and 150 kPa .