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Review Answers with step-by-step examples

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Presentation on theme: "Review Answers with step-by-step examples"— Presentation transcript:

1 Review Answers with step-by-step examples

2 Vocabulary Terms Stoichiometry: Limiting Reagent: Excess Reagent:
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: Actual Yield: Theoretical 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_ CaCl2 + _2_ AgNO­3 _1_ Ca(NO3)2 + _2_ AgCl
How many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate? = 45 g CaCl2 1 mol CaCl2 111 g CaCl2 0.405 mol CaCl2 0.811 mol AgCl 0.405 mol CaCl2 2 mol AgCl 1 mol CaCl2 g AgCl 0.811 mol AgCl 143.5 g AgCl 1 mol AgCl

6 __ C3H8 + _5_ O2­ _4_H2O + _3_CO2 How many moles of O2­ should be supplied to burn 1 mol of C3H8 (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 C3H8 5 mol O2 = 5 mol O2 5 mol O2 6.022 × molecules O2 1 mol O2 = 𝟑.𝟎𝟏𝟏× 𝟏𝟎 𝟐𝟒 Molecules 5 mol O2 32 g O2 1 mol O2 = 160 g O2

7 _4_ Al + _3_ O _2_ Al2O3 Calculate the mass of alumina (Al2O3) 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 Al2O3 = g Al2O3 100 g Al 1 mol Al 27 g Al 3.7 mol Al 2 mol Al2O3 4 mol Al 1.85 mol Al2O3 102 g Al2O3 1 mol Al2O3

8 Review Answers Stoichiometry Practice Problems (Front) 2) 100 g CO2
1) g AgCl 3) g I2 2) 2.21 g 𝐻 2 4) 343 mg H2O or g H2O 3) 34.5 g Na 5) 55 g H2O 4) 292 g Ag Worksheet # 2 Limiting Reactant 5) mol NaBr 1) g Cu / 𝐶𝑢𝑁𝑂 3 is limiting reagent 6) 3.011× Molecules 2) Yes, it is the limiting reagent 7) 3 mol 𝑂 2 3) g CN2H4O / C 𝑂 2 is limiting 8) g Al2O3 4.a) g NaCl / 67.42% yield 9.a) 3214 g H2O or 3.2 kg H2O 4.b) 11.9 g MgCl2 / 82.75% yield 9.b) g Ca(OH) 2 or 13.2 kg Ca(OH) 2 4.c) 6.25 g AlCl3 / 157.6% yield Worksheet # 1 Stoichiometry Practice 5.a) 138 g 𝐻 3 𝑃𝑂 4 1) 770 g K2SO4 5.b) 91.46% yield


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