### Chapter 7.2 Notes

```Chapter 7.2 – Chemical Equations
- chemical reactions can be described two main ways
1. word equation –
write the names of the products and reactants
ex. methane and oxygen yield carbon dioxide and water
2. chemical equation –
uses symbols to show the relationship between the reactants and
products
ex. CH4 + 2O2 → CO2 + 2H2O
Chapter 7.2 – Chemical Equations
- the number in front of a symbol is called a coefficient
- coefficients tell how many of an atom or compound are involved in a
chemical reaction
- the arrow → means yield or get
- it points from the reactants to the products
- chemical equations obey the law of conservation of mass
- because of this they must have equal numbers of atoms of each
element on both sides of the equation
Chapter 7.2 – Chemical Equations
3 steps to writing a balanced chemical equation
1. write a chemical equation by substituting the correct formulas for
the names of the reactants and products
ex. magnesium and oxygen yield magnesium oxide
Mg + O2 → MgO
Chapter 7.2 – Chemical Equations
2. balance the chemical equation one element at a time
- balance elements that only appear once on each side first
- usually balance hydrogen and oxygen last
- only change coefficients, never subscripts
Mg +
O2 →
MgO
Chapter 7.2 – Chemical Equations
3. count the atoms of each element on both sides to be sure the equation
is balanced
2Mg + O2 → 2MgO
reactants
products
Mg
O
Chapter 7.2 – Chemical Equations
Chapter 7.2 – Chemical Equations
Chapter 7.2 – Chemical Equations
law of definite proportions –
a compound always contains the same elements in the same
proportions regardless of how the compound is created or how
much compound is made
- a balanced equation gives a mole ratio –
- the proportion of reactants and products in a chemical equation
ex. 4Al + 3O2 → 2Al2O3
- for every 4 moles of Al need 3 moles of O2 to make 2 moles of Al2O3
Chapter 7.2 – Chemical Equations
- the balanced equation never changes no matter how much Al or O2 we
have
- if I have 8 moles of Al, I need 6 moles of O2 to react with it to give a
mole ratio of 8:6, which is the same as 4:3
- it works for volumes of gases too
2H2O(l) → 2H2(g) + O2(g)
- mole ratio of 2:2:1
- for every 2L of H2 we made, 1L of O2 is also made
Chapter 7.2 – Chemical Equations
- can convert mole ratios to mass as well
4Al + 3O2 → 2Al2O3
- Al molar mass 26.98 g/mol x 4 = 107.92 g Al
- O2 molar mass 32.00 g/mol x 3 = 96.00 g O2
- for Al to react we need 96 g of O2 for every 107.92 g of Al
Chapter 7.2 – Chemical Equations
Chapter 7.2 – Chemical Equations
```