## Chem – Stoichiometry with Energy

HOW DO YOU USE STOICHIOMETRY WITH ENERGY?

You will probably want to have a strong foundation in the previous sections of…….before you learn about this section. If not go back and review those previous sections especially…… Stoichiometry with energy is simply using the different energy notations like you would a mole to mole ratio. To start we should look at a couple of different ways a chemical equation can be presented with energy. Then we can show how to write parts of that equation in a ratio.

The first most common way to see an equation with energy is like below.

200 kJ + N2(g) + 3 H2(g) <—-> 2 NH3(g)

This equation allows us to write ratios to the energy like:

What is the ratio of N2 to the energy?

 1 N2 200 kJ

or

 200 kJ 1 N2

What is the ratio of NH3 to the energy?

 200 kJ 2 NH3

or

 2 NH3 200 kJ

The second most common way to see an equation with energy is like below.

2 C4H10(l) + 13 O2(g) ——> 8 CO2(g) + 10 H2O (g) Δ H = -368 kJ/mol

This equation allows us to write ratios to the energy like:

What is the ratio of O2 to the energy?

 13 O2 368 kJ

or

 368 kJ 13 O2

What is the ratio of CO2 to the energy?

 8 CO2 368 kJ

or

 368 kJ 8 CO2

Notice the negative symbol of the energy does not appear anywhere in the ratios.

VIDEO Stoichiometry with Energy Conversions Demonstrated Example 1: How much energy is required to decompose 4 mol of MnI3?

78 kJ + 2 MnI3(s) ——> 3 I2(g) + Mn(s)

What information does the question supply us with?

What units does the question ask?

How do we set up the problem?

 4 mol MnI3 kJ 1

What is the first conversion?

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 4 mol MnI3 kJ = kJ mol MnI3

What is the next step?

Answer: Fill in the numbers and cross out units

 4 mol MnI3 78 kJ = kJ 2 mol MnI3

Simplify

 4 78 kJ = kJ 2

How do I do the calculations?

Answer: (4 * 78) / 2 = 156

VIDEO Stoichiometry with Energy Conversions Demonstrated Example 2: How many moles of Ca are created when 1400 kJ is released?

6 Ag(s) + Ca3(PO4)2(s) ——> 3 Ca(s) + 2 Ag3PO4(s) Δ H = -382 kJ/mol

What information does the question supply us with?

What units does the question ask?

How do we set up the problem?

 1400 kJ mole Ca 1

What is the first conversion?

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 1400 kJ mole Ca = mole Ca kJ

What is the next step?

Answer: Fill in the numbers and cross out units

 1400 kJ mole Ca = mole Ca kJ

Simplify

 1400 3 mole Ca = mole Ca 382

How do I do the calculations?

Answer: (1400 * 3) / 382 = 11

PRACTICE PROBLEMS: Calculate the energy or moles as needed.

How much energy is required to decompose 5 mol of N2O3?

32 kJ + 2 N2O3(g) ——> 2 N2(g) + 3 O2(g)

How many moles of H2O are created when 97 kJ is released?

2 C2H6(l) + 7 O2(g) ——> 4 CO2(g) + 6 H2O(g) Δ H = -610 kJ/mol

How many moles of CaCl2 are used up when 123 cal is released?

2 Na3PO4(aq) + 3 CaCl2(aq) ——> Ca3(PO4)2(s) + 6 NaCl(aq) + 74 cal