# Stoichiometry Chapter 11. Stoichiometry = the study of quantitative relationships between the amounts of reactants used and products formed by a chemical.

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Stoichiometry Chapter 11

Stoichiometry = the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction –Based on the Law of Conservation of Mass Mass is neither created nor destroyed in a chemical reaction

Law of Conservation of Mass 4Fe + 3O 2  2Fe 2 O 3 Reactants 4 mol Fe x 55.85g Fe = 223.4g Fe 1 mol Fe 3 mol O 2 x 32.00g O 2 = 96.00g O 2 1 mol O 2 Total 319.4 g Products 2 mol Fe 2 O 3 x 159.7g Fe 2 O 3 = 319.4 g Fe 2 O 3 1 mol Fe 2 O 3 Total 319.4 g

Balancing Equations Review Change the coefficients to make the number of atoms of each element equal on both sides of the equation *NEVER change a subscript…doing so changes the identity of the substance Write the coefficients in their lowest possible ratio Check your work

A few things you do NOT do… H 2 + O 2 ---> H 2 O 1.You cannot change a subscript H 2 + O 2 ---> H 2 O 2 2. You cannot place a coefficient in the middle of a formula H 2 + O 2 ---> H 2 2O 3.Make sure that your final set of coefficients are all whole numbers with no common factors other than one 4 H 2 + 2 O 2 ---> 4 H 2 O

Balancing Equations Review Balance this equation NaOH(aq) + CaBr 2 (aq)  Ca(OH) 2 (s) + NaBr(aq) 2NaOH(aq) + CaBr 2 (aq)  Ca(OH) 2 (s) + 2NaBr(aq)

Review: Mole Ratio or (Conversion Factors) Mole Ratio = ratio between the numbers of moles of ANY two substances in a balanced chemical equation

2Al + 3Br 2  2AlBr 3 What mole ratios can be written for this reaction? 2mol Al 3mol Br 2 3molBr 2 2mol AlBr 3 2molAl 2molAlBr 3 2molAlBr 3 2molAl 3mol Br 2 2molAlBr 3 2mol Al 3molBr 2

Helpful Hints ALWAYS make sure the chemical equation is balanced!!! Moles are ALWAYS involved when solving stoichiometric problems The mole ratio is needed to convert from one substance to another in a balanced chemical equation If you are in doubt about how to proceed in solving a problem, GO TO MOLES FIRST!!

Mole to Mole Conversions

How many moles of hydrogen are produced when 0.0400 mole of potassium is used? 2K + 2H 2 0  2KOH + H 2 1.Identify the known… K 2.Identify the unknown…H 2 3.To solve this problem, you need to know how the unknown moles of H 2 are related to the know moles of K 4.The correct ratio should have the moles of unknown in the numerator and the moles of the known should be in the denominator 1mol H 2 2mol K

5.This mole ratio can be used to convert the known number of moles of K to a number of moles of H 2 0.0400mol K x 1mol H 2 = 0.0200mol H 2 2mol K If you put 0.0400mol K into water, 0.0200mol H 2 will be produced

Try one… C 3 H 8 + 5O 2  3CO 2 + 4H 2 O How many moles of CO 2 are produced when 10.0 moles of propane are burned in excess oxygen in a gas grill?

Mole to Mass Conversions

A mole to mass conversion is done when you know the number of moles of a reactant/product and you want to calculate the mass of another reactant/product

Example: Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts vigorously with sodium. 2Na + Cl 2  2NaCl 1.Unknown…NaCl 2.Known…Cl 2 3.Mole ratio…. 2mol NaCl 1 mol Cl 2 4.Multiply the given by the mole ratio 1.25mol Cl 2 x 2mol NaCl = 2.50mol NaCl 1 mol Cl 2

Example: continued… Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts vigorously with sodium. 2Na + Cl 2  2NaCl 5.Multiply the moles of NaCl by the molar mass of NaCl 2.50mol NaCl x 58.45g NaCl = 1mol NaCl = 146g NaCl produced

Try one… Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How many grams of chlorine gas can be obtained from 2.50mol NaCl? 2NaCl  2Na + Cl 2

Mass to Mass Conversions

Example: Determine the mass of H 2 O produced from the decomposition of 25.0g of solid ammonium nitrate. NH 4 NO 3  N 2 O + 2H 2 O 1.Unknown…H 2 O 2.Known… NH 4 NO 3 3.Convert the grams of the given to moles 25.0g NH 4 NO 3 x 1mol NH 4 NO 3 80.06g NH 4 NO 3 = 0.312mol NH 4 NO 3

Example: Determine the mass of H 2 O produced from the decomposition of 25.0g of solid ammonium nitrate. NH 4 NO 3  N 2 O + 2H 2 O 4. Mole ratio… 2mol H 2 O 1mol NH 4 NO 3 5. Multiply by the moles of NH 4 NO 3 previously calculated 0.312mol NH 4 NO 3 x 2mol H 2 O = 0.624mol H 2 O 1mol NH 4 NO 3

Example: Determine the mass of H 2 O produced from the decomposition of 25.0g of solid ammonium nitrate. NH 4 NO 3  N 2 O + 2H 2 O 6.Calculate the mass of H 2 0 using the molar mass conversion factor 0.624mol H 2 O x 18.02g H 2 O = 11.2g H 2 O 1mol H 2 O

Courtesy of Chemistry Matter and Change Glencoe McGraw-Hill Chemistry for Dummies John T. Moore www.karentimberlake.com

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