The stoichiometry of an equation is the ratio of moles of each substance that react or form. You can deduce it from measured masses. Divide each mass by its relative formula mass (Mr) to get moles, then divide all the answers by the smallest number to get the simplest whole-number ratio.
Worked example: 2.4 g of magnesium (Ar 24) reacts with 1.6 g of oxygen (O2, Mr 32) to make 4.0 g of magnesium oxide (MgO, Mr 40). The moles are 0.1 for Mg, 0.05 for O2 and 0.1 for MgO. Dividing by 0.05 gives a ratio of 2 : 1 : 2, so the equation is 2Mg + O2 → 2MgO.
In a real reaction one reactant often runs out first. This is the limiting reactant. When it is used up the reaction stops, and the mass of product depends only on the amount of the limiting reactant. The other reactant is in excess, so some is left over. For example, 4.8 g of magnesium is 0.2 mol and needs 0.1 mol (3.2 g) of oxygen. If only 1.6 g of oxygen is available, the oxygen is limiting and some magnesium is left over. Adding even more magnesium would not make any more magnesium oxide.