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Review Answers with step-by- step examples. Vocabulary Terms Stoichiometry: – the calculation of quantities in chemical equation. Limiting Reagent: –

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Presentation on theme: "Review Answers with step-by- step examples. Vocabulary Terms Stoichiometry: – the calculation of quantities in chemical equation. Limiting Reagent: –"— Presentation transcript:

1 Review Answers with step-by- step examples

2 Vocabulary Terms Stoichiometry: – the calculation of quantities in chemical equation. Limiting Reagent: – the reactant that determines the amount of product in a reaction. Excess Reagent: – A quantity of a reactant that is more than enough to react with a limiting reagent.

3 Percent Yield: – the ratio of the actual yield to the theoretical yield. Actual Yield: – the amount of product formed when a reaction is carried out in the lab. Theoretical Yield: – the calculated amount of product formed during a reaction.

4 Important Things To Remember: Always start with a balanced equation! Remember that you have to go through moles to convert from substance A to substance B.

5 1. _1_ CaCl 2 + _2_ AgNO­ 3 _1_ Ca(NO 3 ) 2 + _2_ AgCl 1.How many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate? = = = 45 g CaCl 2 1 mol CaCl 2 111 g CaCl 2 0.405 mol CaCl 2 2 mol AgCl 1 mol CaCl 2 0.811 mol AgCl143.5 g AgCl 1 mol AgCl 0.405 mol CaCl 2 0.811 mol AgCl 116.38 g AgCl

6 __ C 3 H 8 + _5_ O 2­ _4_H 2 O + _3_CO 2 6.How many moles of O 2­ should be supplied to burn 1 mol of C 3 H 8 (propane) molecules in a camping stove? (write down the equation first; this is a simple Combustion equation and remember that it must be balanced before you continue.) If I had asked for grams instead of molecules, then the last step would be: 1 mol C 3 H 8 5 mol O 2 1 mol C 3 H 8 5 mol O 2 1 mol O 2 5 mol O 2 32 g O 2 1 mol O 2 = 5 mol O 2 = 160 g O 2

7 _4_ Al + _3_ O 2 _2_ Al 2 O 3 8.Calculate the mass of alumina (Al 2 O 3 ) produced when 100 g of aluminum burns in oxygen. (write down the equation first; this is a simple Decomposition equation and remember that it must be balanced before you continue.) = 3.7 mol Al = 1.85 mol Al 2 O 3 = 188.7 g Al 2 O 3 100 g Al1 mol Al 27 g Al 3.7 mol Al2 mol Al 2 O 3 4 mol Al 1.85 mol Al 2 O 3 102 g Al 2 O 3 1 mol Al 2 O 3

8 Review Answers Stoichiometry Practice Problems (Front)2) 100 g CO 2 1) 116.38 g AgCl3) 315.9 g I 2 4) 343 mg H 2 O or 0.343 g H 2 O 3) 34.5 g Na5) 55 g H 2 O 4) 292 g AgWorksheet # 2 Limiting Reactant 5) 0.344 mol NaBr 2) Yes, it is the limiting reagent 8) 188.7 g Al 2 O 3 4.a) 14.61 g NaCl / 67.42% yield 9.a) 3214 g H 2 O or 3.2 kg H 2 O4.b) 11.9 g MgCl 2 / 82.75% yield 9.b) 13214 g Ca(OH) 2 or 13.2 kg Ca(OH) 2 4.c) 6.25 g AlCl 3 / 157.6% yield Worksheet # 1 Stoichiometry Practice 1) 770 g K 2 SO 4 5.b) 91.46% yield


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