The Law of conservation of mass

Back to Combustion! CH4 + 2O2 → CO2 + 2H2O If 575 g of methane gas is combusted with 2300g of oxygen, 1150g of H2O are produced. How much CO2 is produced? CH4 + 2O2 → CO2 + 2H2O Before we can solve this question let’s review an old concept you learned in grade 8!

What is this Law? Nothing is created or destroyed, it only changes form. Therefore, in a chemical reaction, the mass of the reactants is always equal to the mass of the products. Whatever is on the left side of the equation (reactants) must also be on the right side (products). Antoine Laurent Lavoisier (1743-94)

Both sides must be equal Example You combine 40g of Al2O3 with HCl and produce 35g of H2O and 62g of AlCl3, how much HCl did you require? Al2O3 + 6HCl  3H2O + 2AlCl3 40g + xg  35g + 62g Both sides must be equal 40g + x = 35g + 62g x = 35g + 62g – 40g x = 97g – 40g x = 57g

Now we can solve our problem! If 575 g of Methane gas is combusted with 2300g of oxygen, 1150g of H2O are produced. How much CO2 is produced? CH4 + 2O2 → CO2 + 2H2O 575g + 2300g = Xg + 1150g x = 575g + 2300g – 1150g x = 2875g – 1150g x = 1725 g

Now let’s take a look at what is going on at the atomic level! CH4 + O2 → CO2 + H2O What is wrong with this chemical equation? On the reactant side there are 4 hydrogen atoms; on the product side only 2. On the reactant side there are only 2 oxygen atoms; on the product side there are 3.

We cannot forget the law of conservation of mass We cannot forget the law of conservation of mass. Atoms cannot simply disappear or appear. We must balance this equation so that there are equal numbers of each atom on both sides of the equation. CH4 + 2O2 → CO2 + 2H2O By adding coefficients (just like in math) the atoms are now balanced.

CH4 + 2O2 → CO2 + 2H2O How about this combustion reaction? The coefficients indicate the ratios between numbers of molecules: for one methane molecule, two oxygen molecules must react, producing one molecule of carbon dioxide and two water molecules How about this combustion reaction? C6H12O6 + 6 O2 → 6 CO2 + 6 H2O For each molecule of sugar, 6 molecules of oxygen must react, producing 6 molecules of carbon dioxide and 6 molecules of water.

How do I balance an equation? In order to balance an equation you may need to add coefficients. YOU CANNOT ADD SUBSCRIPTS. The coefficient applies to the entire molecule while a subscript applies only to the atom it is attached to.

Let’s Practice counting atoms. How many of each atom are found in the following compounds? 2(AlCl3): Al: ____; Cl: _____ 4Ca(OH)2 : Ca: ___; O: ___; H: ____ 2 6 4 8 8

Steps to follow: Do an inventory of how many atoms of each kind you have on either side of the equation. Start adding coefficients to balance them out (whenever possible, leave H and O for last!!) After each coefficient, be sure to add up each atom again to see how it has changed. You may “mess” up some atoms when fixing others, but that’s okay. Simplify the coefficients in the end if possible.

Examples 2 2 1) CH4 + O2 → CO2 + H2O Now the equation is balanced! C x 1 H x 4 O x 2 H x 2 O x 1 Look at hydrogen: There are 2 in the products and 4 in the reactants. Therefore add a coefficient of 2 before H2O in the products. C x 1 H x 4 O x 2 Look at oxygen. There are 4 in the products and 2 in the reactants. Therefore add a coefficient of 2 before O2 on the reactant side. C x 1 H x 4 O x 4 O x 2 Now the equation is balanced!

2) HCl + Ca(OH)2 → CaCl2 + H2O 2 2 Now the equation is balanced! H x 1 Cl x 1 Ca x 1 O x 2 H x 2 Cl x 2 O x 1 Look at chlorine. There are 2 in the products and 1 on the reactants side. Therefore add a coefficient of 2 before HCl on the reactant side. H x 2 Cl x 2 Ca x 1 O x 2 O x 1 Look at hydrogen. There are 4 on the reactant side and 2 on the product side. Therefore add a coefficient of 2 before H2O on the product side. H x 2 Cl x 2 Ca x 1 O x 2 H x 4 Now the equation is balanced!

3) N2 + O2 → N2O5 2N2 + 5O2 → 2N2O5

Back to steps... Ex: H2 + O2  H2O ≠ H2 + ½O2  H2O 2H2 + O2  2H2O Sometimes you will realize that you need to put a fraction in front of a molecule in order to balance the equation. Once it is balanced, you must then multiply everything by a whole number to get rid of the fraction. There should be no fractions in the final balanced equation. Ex: H2 + O2  H2O ≠ H2 + ½O2  H2O 2H2 + O2  2H2O

More Practice! 4) HCl + O2  Cl2 + H2O 5) KClO3  KCl + O2 4 2 2 2 2 3