The coefficients of a balanced equation give the ratio in which particles react and form. In \mathrm{CH_4 + 2O_2 \longrightarrow CO_2 + 2H_2O}, one methane molecule reacts with two oxygen molecules to give one carbon dioxide molecule and two water molecules.
The ratio does not change with the number of molecules. Taking 6.02 × \mathrm{10^{23}} times as many of each turns the counts into amounts: 1 mol of methane reacts with 2 mol of oxygen to give 1 mol of carbon dioxide and 2 mol of water.
So the ratio of the coefficients is also the ratio of the amounts of substance, 1 : 2 : 1 : 2 here. Every calculation from an equation uses this ratio.
Multiplying each amount by its molar mass gives a mass: 16.0 g of methane reacts with 64.0 g of oxygen to give 44.0 g of carbon dioxide and 36.0 g of water.
The masses are not in the ratio 1 : 2 : 1 : 2, because each substance has its own molar mass. The total, however, is 80.0 g on both sides, since the atoms are only rearranged.
For gases the amounts can also be turned into volumes with the molar volume, 22.4 L/mol at 0 °C and 1.013 × \mathrm{10^5} Pa: 22.4 L of methane reacts with 44.8 L of oxygen to give 22.4 L of carbon dioxide, while the water is a liquid.
To find the mass of carbon dioxide formed when 8.00 g of methane burns, change the mass into an amount, use the ratio of the coefficients, and change the amount back into a mass.
The methane is 8.00 g ÷ 16.0 g/mol = 0.500 mol. Methane and carbon dioxide have the same coefficient, so 0.500 mol of carbon dioxide forms, which is 0.500 mol × 44.0 g/mol = 22.0 g.
The oxygen used follows in the same way: 0.500 mol × 2 = 1.00 mol, or 32.0 g. The water formed is also 1.00 mol, 18.0 g, and the masses balance: 8.00 g + 32.0 g = 22.0 g + 18.0 g.
Sodium hydrogencarbonate, the baking soda that makes cakes rise, breaks down when heated: \mathrm{2NaHCO_3 \longrightarrow Na_2CO_3 + H_2O + CO_2}. The carbon dioxide can be collected over water in an upturned measuring cylinder.
The molar mass of \mathrm{NaHCO_3} is 84.0 g/mol, so 1.68 g is 0.0200 mol. The coefficients 2 and 1 give half as much carbon dioxide, 0.0100 mol.
At 0 °C and 1.013 × \mathrm{10^5} Pa this takes up 0.0100 mol × 22.4 L/mol = 0.224 L, or 224 mL.
The white solid left in the tube is 0.0100 mol of sodium carbonate, 1.06 g. The 0.62 g lost is the water and carbon dioxide that left the tube.
In a solution the amount comes from the concentration and the volume. Hydrogen peroxide solution gives off oxygen when a little manganese(IV) oxide is added: \mathrm{2H_2O_2 \longrightarrow 2H_2O + O_2}. The manganese(IV) oxide speeds up the reaction and is not used up.
Take 10.0 mL of 1.00 mol/L hydrogen peroxide solution. It contains 1.00 mol/L × 0.0100 L = 0.0100 mol of \mathrm{H_2O_2}, and the coefficients 2 and 1 give 0.00500 mol of oxygen.
At 0 °C and 1.013 × \mathrm{10^5} Pa this is 0.00500 mol × 22.4 L/mol = 0.112 L, or 112 mL. Twice the volume of solution gives twice the oxygen, while more manganese(IV) oxide only makes the oxygen come faster.