#  Balance the following equation.  Fe + O 2  Fe 2 O 3  4Fe + 3O 2  2Fe 2 O 3  This means that when we combine four atoms of iron with three molecules.

## Presentation on theme: " Balance the following equation.  Fe + O 2  Fe 2 O 3  4Fe + 3O 2  2Fe 2 O 3  This means that when we combine four atoms of iron with three molecules."— Presentation transcript:

 Balance the following equation.  Fe + O 2  Fe 2 O 3  4Fe + 3O 2  2Fe 2 O 3  This means that when we combine four atoms of iron with three molecules of oxygen gas we will get to molecules of iron oxide.  It also means that if we combine four MOLES of iron with three MOLES of oxygen we can make 2 MOLES iron oxide.

 Balance the following equation.  C 3 H 8 + O 2  CO 2 + H 2 O  C 3 H 8 + 5O 2  3CO 2 + 4H 2 O  How many grams of C 3 H 8 do we start with?  44.1 g  How many grams of O 2 do we start with?  160g  How much do our products weight?  204.1g

 We can use the coefficients of balanced chemical equations to set up ratios.  Example:  2K + Br 2  2KBr  We can write six mole ratios for this equation.

 What are all the possible mole ratios we can write for the following equation?  2KClO 3  2KCl + 3O 2

 Determine all possible mole ratios for the following balanced chemical equations.  4Al + 3O 2  2Al 2 O 3  3Fe + 4H 2 O  Fe 2 O 4 + 4H 2  2HgO +  2Hg + O 2

 The first step to solving a stoichiometric problem is having a balanced equation.  K + H 2 O  KOH + H 2  2K + 2H 2 O  2KOH + H 2  Then we can use mole ratios to solve problems.  If 2 moles of Potassium makes 1 mole of H 2 then how many moles of H 2 will one mole of Potassium make?

 C 3 H 8 + 5O 2  3CO 2 + 4H 2 O  How many molecules of CO 2 are produced when 10.0 moles of C 3 H 8 is burned with an excess amount of O 2 ?

 Fe + O 2  Fe 2 O 3  How many moles of oxygen are required to completely react with 6 moles of iron?  4Fe + 3O 2  2Fe 2 O 3  4.5 mol O 2  Cu + AgNO 3  Cu(NO 3 ) 2 + Ag  How many moles of Cu(NO 3 ) 2 can be produced from reacting 0.035 moles of AgNO 3 ?  Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag  0.0713 moles of Cu(NO 3 ) 2

 Na + H 2 O  NaOH + H 2  How many grams of NaOH can be produced from the reaction of 22.98 grams of Na?  2Na + 2H 2 O  2NaOH + H 2  39.98 g of NaOH  How many grams of H 2 can be produced from the reaction of 24 g of H 2 O?  1.33 g H 2

 The reaction of copper (II) chloride and aluminum is a single replacement reaction.  Al + CuCl 3  AlCl 3 + Cu  2Al + 3CuCl 3  2AlCl 3 + 3Cu  If our piece of Al initially weights 1.56g and after the reaction it weights 1.24 g predict the amount of copper in grams that we produced.

 Limiting reactants, like the name says, limits the reaction.  In any reaction the limiting reactant will be completely used up and then the reaction will stop.  Limiting reactants will also limit the amount of product we can make.

 How can you calculate the amount of products when one reactant is limiting?  S 8 + 4Cl 2  4S 2 Cl 2  If 200.0 g of S 8 reacts with 100.0 g of Cl 2 what is the mass of S 2 Cl 2 produced?  When the mass of both reactants is given the first step is to determine which one is the limiting reactant.

 S 8 + 4Cl 2  4S 2 Cl 2  100.0 g Cl 2 /70.91 (g/mol) Cl 2 = 1.410 mol Cl 2  200.0 g S 8 /256.5 (g/mol) S 8 = 0.7797 mol S 8  The next step is determining which reactant will run out first.  The coefficients from the balanced equation tell us that 4 moles of Cl 2 are needed to react with one mole of S 8.  If we have 0.7797 moles of S 8 how many moles of Cl 2 do we need to completely react all of the S 8 ?  4 x 0.7797 = 3.12 mol Cl 2  But we only have 1.410 moles of Cl 2 so Cl 2 is our limiting reactant.

 16Ag + S 8  8Ag 2 S  a.) If 4.00 g of Ag and 4.00 g of S 8 react, which is the limiting reactant?  Ag  b.) How much of the excess reactant is left over?  0.0133 mol S 8 left over.  c.) How many moles of Ag 2 S will be produced?  0.0184 moles

 2Al + 3I 2  2AlI 3  a.) If 54.0 g of Al reacts with 50.8 g of I 2, which is the limiting reactant? I2I2  b.) How much of the excess reactant will be left over?  0.6 moles Al left over  c.) How much product will be produced?  1.4 moles of AlI 3 produced

 It’s all about the balanced equation and mole ratios.  If one reactant is “in excess” you don’t need to worry about it.  If you’re given moles you can use the mole ratio right away. If you’re given grams you have to convert to moles first.  Think about what the question is asking.  When in doubt follow the unit conversion process.  When given amounts of both reactants you must figure out which is the limiting reactant.

 Set problems up and then use calculator  mass before = mass after

 N 2 + 3H 2  2NH 3  How many moles of NH 3 are produced when 25.0 g of N 2 react with excess H 2 ?  2KClO 3  2KCl + 3O 2  If 5.0 g of KClO 3 decomposes how many grams of O 2 will be produced?  How many grams of KCl will be produced?  Zn + 2HCl  ZnCl 2 + H 2  How many moles of hydrogen are produced when 2.5 g of Zn react with excess hydrochloric acid.

 N 2 + 3H 2  2NH 3  How many grams of NH 3 can be produced from the reaction between 28 g of N 2 and 25 g of H 2 ?  How much of the excess reactant is left over?  Mg + 2HCl  MgCl 2 + H 2  How many grams of hydrogen will be produced when 50 g of Mg react with 75 g of HCl?  How much of the excess reactant is leftover?

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