Chapter 9 Stoichiometry 9.2 chemical calculations

Chapter 9Stoichiometry

9.2 chemical calculations

Things you will learn

• You will be able to construct mole ratios from balanced chemical equations and be able to apply these in mole-mole calculations.

• You will be able to calculate stoichiometric quantities (mass, volume, representative particles, and moles) from balanced equations.

Baking soda experiment

• Baking soda decomposes with heat to form sodium carbonate, carbon dioxide, and water

• The skeleton equation is:• NaHCO3 Na2CO3 + CO2 + H2O

Baking soda experiment

• Baking soda decomposes with heat to form sodium carbonate, carbon dioxide, and water

• The balanced equation is:• 2NaHCO3 Na2CO3 + CO2 + H2O

2 Na 22 C 26 O 62 H 2

What we want to find out is how much baking soda will we have to heat in order to fill a liter Coke bottle half-full.Since we have a balanced equation, we can use the molar road map to find out just how much we should use.We will start by putting the given quantity on the left and the units of what we want to find on the right, and filling in the middle with conversion factors.

Given quantity

Wanted quantity

Mole-mole ratio from balanced equation

.5 L CO2

1 mol CO222.4 L CO2

2 mol NaHCO31 mol CO2

84 g NaHCO31 mol NaHCO3

_ g NaHCO3

Given quantity

Wanted quantity

Ratio from balanced equation

3.75 g NaHCO3

So if we put 3.75 grams of NaHCO3 in a test tube and heated it, we should get ½ liter of CO2 given off.

What else is given off that might mess up our calculations?

2NaHCO3 Na2CO3 + CO2 + H2O

So if we put 3.75 grams of NaHCO3 in a test tube and heated it, we should generate ½ liter of CO2.

What else is given off that might mess up our calculations?

2NaHCO3 Na2CO3 + CO2 + H2O

These are solids These are BOTH gases!

Mole-mole calculations

• 4Al(s) + 3O2(g) 2Al2O3(s)

• Using this balanced equation, what are the possible mole ratios we can use as conversion factors?

• Remember, conversion factors are always equal to 1 and so may be used upside down, with the numerator in the denominator, and vice versa

Mole-mole calculations

• 4Al(s) + 3O2(g) 2Al2O3(s)

• Using this balanced equation, what are the possible mole ratios we can use as conversion factors?

4 moles Al3 moles O2

3 moles O2

4 moles Al2 moles Al2O3

3 moles O2

3 moles O2

2 moles Al2O3

2 moles Al2O3

4 moles Al

4 moles Al2 moles Al2O3

We use mole-mole conversions (ratios from balanced equations) to solve problems like:How many moles of oxygen are required to react completely with 14.8 mole of aluminum?We always work from our known (given) quantity towards our unknown (wanted) quantity.

14.8 mol Al3 mol O2 4 mol Al

_ mol O2

givenmole-mole ratio unknown

Mole-mole calculations

• 4Al(s) + 3O2(g) 2Al2O3(s)

• How many moles of Al2O3(s) would we get if we only used 1 mole of Al(s)?

1 mol Al2 mol Al2O3

4 mol Al_ mol Al2O3

givenmole-mole ratio unknown

Mole-mole calculations

• 4Al(s) + 3O2(g) 2Al2O3(s)

• How many moles of Al2O3(s) would we get if we only used 1 mole of O2(g)?

1 mol O22 mol Al2O3

3 mol O2_ mol Al2O3

given mole ratio

unknown

Mole-mole calculations

• 4Al(s) + 3O2(g) 2Al2O3(s)

• How many moles of O2(g) would it take to make 5 moles of Al2O3(s) ?

5 mol Al2O33 mol O2

2 mol Al2O3_ mol O2

given mole ratio

unknown

Mole-mole calculations

• 4Al(s) + 3O2(g) 2Al2O3(s)

• How many moles of Al(s) would it take to make 5 moles of Al2O3(s) ?

14.8 mol Al3 mol O2 4 mol Al

_ mol O2

given mole ratio

unknown

Mass-mass calculations

• The number of moles of a reactant or product can be gotten directly from the coefficients of a balanced equation.

• The amount in grams must be gotten from the molar mass of each. This adds a couple more conversion factors, but they’re easy to deal with.

Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of

hydrogen with an excess of nitrogen

• N2(g) + H2(g) NH3(g) (skeleton equation)

Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of

hydrogen with an excess of nitrogen

• N2(g) + 3H2(g) 2NH3(g) (balanced equation)

Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of

hydrogen with an excess of nitrogen

• N2(g) + 3H2(g) 2NH3(g)

5.4 g H2

_ g NH3

given Gram-mole conversion

Mole-mole conversion

Gram-mole conversion

unknown

Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of

hydrogen with an excess of nitrogen

• N2(g) + 3H2(g) 2NH3(g)

5.4 g H2

1 mol H22 g H2

2 mol NH33 mol H2

17 g NH31 mol NH3

_ g NH3

given Gram-mole conversion

Mole-mole conversion

Gram-mole conversion

unknown

Calculate the number of grams of ammonia produced from the reaction of 5.4 grams of

hydrogen with an excess of nitrogen

• N2(g) + 3H2(g) 2NH3(g)

5.4 g H2

1 mol H22 g H2

2 mol NH33 mol H2

17 g NH31 mol NH3

30.6 g NH3

given Gram-mole conversion

Mole-mole conversion

Gram-mole conversion

unknown

Mole-mole conversionsAmmonia burns in oxygen forming nitrogen

oxide and water. The skeleton formula is: NH3(g) + O2(g) NO(g) + H2O(g)

Balance this equation and determine how many moles of ammonia are consumed if 1.35

moles of oxygen are used.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

4 N 412 H 1210 O 10

1.35 mol O24 mol NH35 mol O2

_mol NH3

Mass-mole conversionsPhosphorus reacts with chlorine to form phosphorus pentachloride. The skeleton formula is: P(s) + Cl2(g) PCl5(s)

Balance this equation and determine how many moles of chlorine are consumed if 5.50 grams of PCl5 are produced.

2P(s) + 5Cl2(g) 2PCl5(s)

2 P 210 Cl 10

5.50 g PCl51 mol PCl5208 g PCl5

_mol Cl25 mol Cl2

2 mol PCl5