Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions. Stoichiometry is all about the numbers. All reactions are dependent on how much stuff you have. It helps you figure out how much of a compound you will need, or maybe how much you started with.
When you're doing problems in stoichiometry, you might look at:
1) Mass of reactants
2) Mass of products
3) Chemical equations
4) Molecular weights of either the reactant or product
5) Formulas of various compounds
Trick:
Mol Box:
grams x grams y
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moles x ------------------------> moles y
Grams to Moles / Moles to Grams: molecular mass
Moles to Moles: molar ration from the balance chemical equation
Examples:
1) In the formation of carbon dioxide from carbon monoxide and oxygen, how many moles of carbon monoxide are needed to react completely with 7.0 moles of oxygen gas?
2 CO (g) + O2(g) → 2 CO2 (g)
* We have moles of x and we need moles of y
7.0 mol O2 x 2 mol CO = 14 moles of CO
1 mol O2
2) How many grams of oxygen are produced by the decomposition of 1 mole of potassium chlorate, KClO3?
2 KClO3 → 2 KCl + 3 O2
*We have moles of x and we need grams of y
1 mol KCLO3 x 3 mol O2 x 32g O2 = 48g O2
2 mol KCLO3 1 mol O2
When you're doing problems in stoichiometry, you might look at:
1) Mass of reactants
2) Mass of products
3) Chemical equations
4) Molecular weights of either the reactant or product
5) Formulas of various compounds
Trick:
Mol Box:
grams x grams y
| |
| |
| |
moles x ------------------------> moles y
Grams to Moles / Moles to Grams: molecular mass
Moles to Moles: molar ration from the balance chemical equation
Examples:
1) In the formation of carbon dioxide from carbon monoxide and oxygen, how many moles of carbon monoxide are needed to react completely with 7.0 moles of oxygen gas?
2 CO (g) + O2(g) → 2 CO2 (g)
* We have moles of x and we need moles of y
7.0 mol O2 x 2 mol CO = 14 moles of CO
1 mol O2
2) How many grams of oxygen are produced by the decomposition of 1 mole of potassium chlorate, KClO3?
2 KClO3 → 2 KCl + 3 O2
*We have moles of x and we need grams of y
1 mol KCLO3 x 3 mol O2 x 32g O2 = 48g O2
2 mol KCLO3 1 mol O2