A quick Google will answer your question, research.
That said, the reaction order will be related to the stoichiometry of the chemical equation only for very simple reactions, e.g. X+Y —>Z where X and Y collide and form molecule Z. If you add more of either X or Y, the collisions are more frequent and the reaction is faster. This gives us the rate equation d[Z]/dt=k[X][Y]. The reaction order is the sum of the indices above the square brackets. Note, we could write the previous equation as d[Z]/dt=k[X]1[Y]1 so, as 1+1=2, this reaction is second order, as is 2X —>Y, where d[Y]/dt=k[X]2.
Now let's suppose that molecule Y in the first reaction is water and we're conducting the reaction in water, for example, CO2 + H2O —> H2CO3
Because water is the solvent, the CO2 molecules are always colliding with water molecules — adding more water won't increase the rate of production of H2CO3. However, adding more CO2 will increase the reaction rate so the reaction looks like it's first order with respect to CO2. This reaction will also appear to be zero order with respect to water, (as we said, adding more water will not affect the rate of the reaction). in other words, it's pseudo-first order: d[H2CO3]/dt = k[CO2].
A zero order reaction will not depend on the concentration of the chemical at all. The rate of a thermal decomposition, for instance, will often depend on the temperature of the system rather than the concentration of the reactant.
In addition, many reactions are more complex than molecules simply knocking into one another and these may have third order kinetics or even fractional orders.
For all of these reasons, the order of a reaction is only related to stoichiometry for very simple reactions and (generally) only for first or second order reactions.