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? I2I2 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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