Scientific Understanding

**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 + N_{2(g)} + 3 H_{2(g)} <—-> 2 NH_{3(g)}

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

What is the ratio of N_{2} to the energy?

1 N_{2} |

200 kJ |

or

200 kJ |

1 N_{2} |

What is the ratio of NH_{3} to the energy?

200 kJ |

2 NH_{3 } |

or

2 NH_{3} |

200 kJ |

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

2 C_{4}H_{10(l) } + 13 O_{2(g) } ——> 8 CO_{2(g) } + 10 H_{2}O_{ (g) }**Δ** H = -368 kJ/mol

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

What is the ratio of O_{2} to the energy?

13 O_{2} |

368 kJ |

or

368 kJ |

13 O_{2 } |

What is the ratio of CO_{2} to the energy?

8 CO_{2} |

368 kJ |

or

368 kJ |

8 CO_{2 } |

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 MnI_{3}?

78 kJ + 2 MnI_{3(s)} ——> 3 I_{2(g)} + Mn_{(s)}

What information does the question supply us with?

Answer: 4 mol MnI_{3}

What units does the question ask?

Answer: kJ

How do we set up the problem?

Answer:

4 mol MnI_{3} |
kJ | |

1 |

What is the first conversion?

Answer: mole to kJ ratio

How do I put that in?

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

4 mol MnI_{3} |
kJ = | kJ |

mol MnI_{3} |

What is the next step?

Answer: Fill in the numbers and cross out units

4 mol MnI_{3} |
78 kJ = | kJ |

2 mol MnI_{3} |

Simplify

4 | 78 kJ = | kJ |

2 |

How do I do the calculations?

Answer: (4 * 78) / 2 = 156

What is the complete answer?

COMPLETE ANSWER: 156 kJ

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

6 Ag_{(s)} + Ca_{3}(PO_{4})_{2(s)} ——> 3 Ca_{(s)} + 2 Ag_{3}PO_{4(s) }**Δ** H = -382 kJ/mol

What information does the question supply us with?

Answer: 1400kJ

What units does the question ask?

Answer: mole Ca

How do we set up the problem?

Answer:

1400 kJ | mole Ca | |

1 |

What is the first conversion?

Answer: mole to kJ ratio

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

What is the complete answer?

COMPLETE ANSWER: 11 mole Ca

**PRACTICE PROBLEMS**: Calculate the energy or moles as needed.

How much energy is required to decompose 5 mol of N_{2}O_{3}?

32 kJ + 2 N_{2}O_{3(g)} ——> 2 N_{2(g)} + 3 O_{2(g)}

Answer: 80.0 kJ

How many moles of H_{2}O are created when 97 kJ is released?

2 C_{2}H_{6(l)} + 7 O_{2(g)} ——> 4 CO_{2(g)} + 6 H_{2}O_{(g) }**Δ** H = -610 kJ/mol

Answer: 0.954 mole H_{2}O

How many moles of CaCl_{2} are used up when 123 cal is released?

2 Na_{3}PO_{4(aq)} + 3 CaCl_{2(aq)} ——> Ca_{3}(PO_{4})_{2(s)} + 6 NaCl_{(aq)} + 74 cal

Answer: 4.99 mole CaCl_{2}

How much energy is required to product 10.5 mole of C_{6}H_{12}O_{6}?

6 C_{4(s)} + 12 O_{2(g)} + 24 H_{2(g)} ——> 4 C_{6}H_{12}O_{6(s) }**Δ** H = 78000 J/mol

Answer: 2.05 * 10^{5} J