Stoichiometry is a very important part of chemistry. It is used in designing experiments with chemical reactions. Before learning about stoichiometry, brush up on moles and balancing chemical reactions as they are both important to understanding stoichiometry.
The number in front of reactants and products in a chemical equations says the number of molecules of each reactant you need or of each product that will be produced. In the methane combustion reaction above, for example, you need 1 molecule of methane for every 2 molecules of oxygen. The result will be 1 molecule of carbon dioxide and 2 molecules of water.
When Chemists design chemical reactions, it is nearly impossible to isolate an exact number of molecules of each atom. This is why they use moles. 1 mole is always 6.022X10^23 molecules. This means that you can replace the proportionality of molecules with the same proportionality of moles. In the reaction above, that means that for every mole of methane, you need 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
Stoichiometry is finding the right quantity of a compound to make the reaction works. This involves finding the right amount of moles to complete an equation, then determining how much mass of that substance would make that amount of moles. Let's try an example.
If this section is confusing, check out our page on balancing reactions.
The limiting reactant is the reactant that there isn't enough of to react with all of the other reactant.
The other reactant still reacts with the limiting reactant; however, only some of it will since there won't be enough of the limiting reactant to react with.
When taksed with finding the limiting reactant, you will be given the mass of both reactants. You convert those masses to moles. Divide both of the moles of the reactants by the number of molecules needed to complete the reaction (the number in front of the chemical formula in the reaction equation). The reactant with the lower value is the limiting reactant.
You can do a stoichiometry problem with a limiting reactant. All you have to do is follow the example above, but start with the limiting reactant and use its mole value to find the mass of the products.
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