UNIT VII Stoichiome-tree Lesson 1. VII.1 C OEFFICIENTS IN A R EACTION E QUATION : Stoichiometry : The relationship between the amount of reactants used.

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Presentation transcript:

UNIT VII Stoichiome-tree Lesson 1

VII.1 C OEFFICIENTS IN A R EACTION E QUATION : Stoichiometry : The relationship between the amount of reactants used in a chemical reaction and the amounts of products produced by the reaction

VII.1 C OEFFICIENTS IN A R EACTION E QUATION : 2Mg + O 2 → 2MgO The coefficient 2, means two magnesium atoms react with one oxygen molecule to produce two molecules of magnesium oxide if you double the amounts of Mg and O 2, what will happen to the amount of MgO??? if you use 10 times the amount of reactants, what will happen to the amount of MgO???

VII.1 C OEFFICIENTS IN A R EACTION E QUATION : *the balanced equation describes the ratio in which the substances combine In terms of MOLES: 2(6.02 x )Mg + (6.02 x )O 2 → 2(6.02 x )MgO 2 mol Mg + 1 mol O 2 → 2 mol MgO The ratio

VII.1 C OEFFICIENTS IN A R EACTION E QUATION : Ex: Consider the reaction equation 2Na + Cl 2 → 2 NaCl a.)How many atoms of Na are required to react with 10 molecules of Cl 2 ? b.) How many moles of NaCl are produced when 15 mol of Na are reacted? p. 124 #1-5

VII.1 C OEFFICIENTS IN A R EACTION E QUATION : Consider the reaction equation:

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES * A chemical equation is written in terms of moles of reactants and products To predict the moles of each substance, you can use the mole ratio Stoichiometry calculations allow you to calculate the amount of CHEMICAL #1 involved in a reaction, based on the amount of CHEMICAL #2

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES From UNIT 5 ( The Mole Unit), we know how to get from L to moles, g to moles, molecules to moles etc.... MOLES are the way to go from CHEMICAL #1 to CHEMICAL #2MOLES ofCHEMICAL #1 #2 ALWAYS convert your values into moles ---- then cross the mole bridge!!

I N OTHER WORDS … ANYTIME AT ALL, you are asked to calculate a mass, # of molecules, or a volume between TWO chemicals, each on one side of the equation, you MUST change it first into moles, second use the mole ratio between these two chemicals you have, then third change back to whatever units the question wants! units u have  moles  mole ratio  units you want

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES Ex #1: How many moles of sodium metal would be needed to react with chlorine gas and make 737 g of sodium chloride ? 2Na + Cl 2 → 2NaCl

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES Ex: #2 How many grams of potassium chloride, KCl, are produced by decomposing118 g of potassium chlorate, KClO 3 ?

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES TRY C 3 H 8(g) + 5O 2(g) → 3CO 2(g) + 4H 2 O (l) A sample of propane is burned, what mass of H 2 O (l) is produced if the reaction also produces 50.0L of CO 2(g) at STP?

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES Ex: #3 Consider the reaction… 4NH 3 + 5O 2 → 6H 2 O + 4NO How many molecules of NH 3 are required to react with 3.00 mol of O 2(g) ?

VII.2 S TOICHIOMETRY C ALCULATIONS INVOLVING M OLES, M ASS, AND G AS V OLUME AND M OLECULES TRY: 2NH 3(aq) + NaOCl (aq) → N 2 H 4(aq) + NaCl(aq) + H 2 O(l) If 1.25 x 10 4 kg of N 2 H 4 is produced, how many litres of ammonia gas, at STP, is required in the reaction?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Molar Concentration: c = n/V Only use 22.4L/mol if it is a gas at STP

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Ex: # 1 If copper wire is put in a solution of silver nitrate, how many grams of copper will react to completely replace silver from 208 mL of 0.100M solution of silver nitrate, AgNO 3 ?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Ex: #2 For the reaction: Zn + 2HCl → ZnCl 2 + H 2 a) What volume of 3.00 M HCl is required to react with 12.35g of zinc?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Ex: #2 (continued) For the reaction: Zn + 2HCl → ZnCl 2 + H 2 b) How many moles of hydrogen are produced when 12.35g of zinc are reacted with the correct amount of HCl?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Ex: #3 What volume of CO 2(g) at STP is produced if 2.30 L of 0.5 M HCl reacts with an excess of CaCO 3 ?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION TITRATIONS Definition: a process used to determine the unknown concentration of a solution (SAMPLE) by reacting it with a measured amount of a solution with known concentration (STANDARD) until a desired equivalence point is reached. Usually, it is an acid-base titration.

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Equivalence Point = Stoichiometric Point Point in titration where the ratio of the moles of each species involved exactly equals the ratio of the coefficients of the species in the balanced reaction equation Ratio of COEFFICIENTS = Ratio of MOLES this point the indicator changes colour

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Ex: #1 H 3 PO 4 + 2KOH → K 2 HPO 4 + 2H 2 O What volume of M KOH is required to react with 125mL of M H 3 PO 4 in order to produce a solution of K 2 HPO 4 ?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION Ex: #2 H 3 PO 4 + 2KOH → K 2 HPO 4 + 2H 2 O If 19.8 mL of H 3 PO 4 with an unknown molarity react with 25.0 mL of M KOH, according to the above reaction, what is the molarity of the H 3 PO 4 ?

VII.3 S TOICHIOMETRY C ALCULATIONS I NVOLVING M OLAR C ONCENTRATION TRY H 2 SO 4 + 2NaOH → Na 2 SO 4 + 2H 2 O If 20.8 mL of H 2 SO 4 with an unknown molarity react with 23.0 mL of M NaOH, according to the above reaction, what is the molarity of the H 2 SO 4 ?

HOMEWORK Meaning of coefficients: p. 124 # 2, 3. Stoichiometry: p. 127 # 6,7,8,9,12 Molar Concentrations: p. 131 #