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- and how to balance them

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1 - and how to balance them
Chemical equations - and how to balance them

2 There are several ways to represent a chemical reaction.
in a long sentence : Iron, a solid, reacts with oxygen, a gas, to form iron oxide, a solid. this tells a chemist very little, and shows nothing about amounts of substance a word equation Iron + oxygen  iron oxide Shorter, but still no information on quantities. using symbols – the symbol equation 4Fe + 3O2  2Fe2O3 Much better – quicker to write and gives a ratio of all reactants (on the left) and products (on the right). Can be refined further with state symbols to show the state of each substance: 4Fe(s) + 3O2(g)  2Fe2O3(s) (s) = solid (l) = liquid (g) = gas (aq) = aqueous solution (solution in water)

3 Writing symbol equations
e.g. 2Mg(s) + O2(g)  2MgO(s) This is known as a balanced symbol equation as it applies the law of conservation of matter: Matter can neither be created nor destroyed. The same number of atoms that go into a reaction must come out. The change that occurs is in the arrangement of the atoms: bonds are being broken (between atoms of Mg, and between atom of O) the atoms rearranged (Mg meets O) new bonds are formed (between Mg atoms and O atoms) making new substances (MgO is formed)

4 A stepwise process: Write out formulae of the reactants (on the left) and the products (on the right) and place an arrow (going right) in between. Important to learn some basic formulae, especially of elements that exist as molecules (the molecular elements). These are: H2, N2, O2, and the halogens: F2, Cl2, Br2, I2. All other elements can be considered to be single atoms. Count each type of atom on either side of the equation. Place numbers in front of formulae to multiply the atoms present – this is the tricky bit!! NB do not change the formulae!! Add state symbols if known.

5 Example – magnesium burning in oxygen
the formulae: Mg + O2  MgO count the atoms: 1x Mg 2xO 1xMg + 1xO easy to see that one more O is needed on the product side, place a 2 in front of MgO, giving 2MgO: Mg + O2  2MgO recount atoms: 1xMg 2xO 2xMg + 2xO Mg is now unbalanced – place a 2 in front of the Mg, giving 2Mg 2Mg + O2  2MgO balanced add state symbols 2Mg(s) + O2(g)  2MgO(s)

6 Example – methane burning in oxygen
the formulae: CH4 + O2  CO2 + H2O C H O C O O H 1xC 4xH 2xO 1xC 2xH 3xO C O O H H H H C O O O H H Easy to double Hs first, so 2H2O CH4 + O2  CO2 + 2H2O C H O C O O H O H Recounting atoms 1xC 4xH 2xO 1xC 4xH 4xO

7 Just need to double O2, so 2O2, left to last, so nothing else is altered.
CH4 + 2O2  CO2 + 2H2O C H O C O O H O

8 What needs to be done? Think before you click!!
Harder, longer (slightly different layout): acid + carbonate The formulae: HNO3 + Na2CO3  NaNO3 + CO2 + H2O As NO3- is on both sides it can be counted as one unit, like an atom. CO32- changes, so C and O need counting. H 1  2 NO3- 1  1 Na 2  1 C 1  1 O 3  3 What needs to be done? Think before you click!! Need to double Hs (2HNO3) on the left and Na (2NaNO3) on the right – note that this doubles the NO3- on both sides!! Very convenient.

9 Recount the atoms (think first!!): H 2  2 NO3- 2  2 Na 2  2 C 1  1
2HNO3 + Na2CO3  2NaNO3 + CO2 + H2O Recount the atoms (think first!!): H 2  2 NO3- 2  2 Na 2  2 C 1  1 O 3  3 all balanced State symbols: 2HNO3(aq) + Na2CO3(aq)  2NaNO3(aq) + CO2(g) + H2O(l)


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