Download presentation

1
**The BCA Method in Stoichiometry**

Larry Dukerich Modeling Instruction Program Arizona State University

2
**Typical Approach to Stoichiometry**

Very algorithmic grams A -->moles A-->moles B-->grams B Fosters plug-n-chug solution Disconnected from balanced equation Especially poor for limiting reactant problems

3
BCA Approach Stresses mole relationships based on coefficients in balanced chemical equation Sets up equilibrium calculations later (ICE tables) Clearly shows limiting reactants

4
**Balanced Chemical Equations**

Atoms are conserved in chemical reactions: balanced chemical equations. Reactants --> Products Ratio of particles: mole ratios through coefficients Ex. 2 Mg + 1 O2 --> 2MgO two moles of magnesium burn with one mole of oxygen gas to produce two moles of MgO 2 moles Mg: 1 mole O2: 2 moles MgO

5
**Making Predictions Using mole ratios we can identify:**

How much is needed (before) How much can be made (after) To make predictions we need also to consider the changes that occur during the reaction. BCA: Before-Change-After chart to keep track of what happens to particle/mole ratios during a chemical reaction

6
**Emphasis on balanced equation**

Step 1- Balance the equation Hydrogen sulfide gas, which smells like rotten eggs, burns in air to produce sulfur dioxide and water. How many moles of oxygen gas would be needed to completely burn 2.4 moles of hydrogen sulfide? H2S O2 ----> 2 SO H2O Before: Change After

7
**Focus on mole relationships**

Step 2: fill in the before line 2 H2S O2 ----> 2 SO H2O Before: xs Change After Assume more than enough O2 to react

8
**Focus on mole relationships**

Step 3: use ratio of coefficients to determine change 2 H2S O2 ----> 2 SO H2O Before: xs Change –2.4 – After Reactants are consumed (-), products accumulate (+)

9
**Emphasize that change and final state are not equivalent**

Step 4: Complete the table 2 H2S O2 ----> 2 SO H2O Before: xs Change –2.4 – After xs

10
**Complete calculations on the side**

In this case, desired answer is in moles If mass is required, convert moles to grams in the usual way

11
Mole Ratios: #1 Lead will react with hydrochloric acid to produce lead chloride (PbCl2) and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: Before: Change: ____________________________________________________ After:

12
Equation: Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + __ HCl --> __PbCl2 + __H2 Before: Change: ____________________________________________________ After:

13
Balance & Before: Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + _2_ HCl --> __PbCl2 + __H2 Before: moles excess 0 moles 0 moles Change: ____________________________________________________ After:

14
Change Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + _2_ HCl --> __PbCl2 + __H2 Before: moles excess 0 moles 0 moles Change: -4 moles -8moles mole + 4 mole ____________________________________________________ After:

15
After Lead will react with hydrochloric acid to produce lead chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 4.0 mole of lead? Equation: __Pb + _2_ HCl --> __PbCl2 + __H2 Before: moles excess 0 moles 0 moles Change: -4 moles -8 moles mole + 4 mole ____________________________________________________ After: 0 mole excess 4 moles 4 moles You Can Convert: moles to grams (multiply moles x molar mass (g/mol))

16
Mole Rations: #2 How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH H2O --> Ca(OH) H2 Before Change ___________________________________________ After

17
Before: How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH H2O --> Ca(OH) H2 Before moles excess moles 0 moles Change ___________________________________________ After

18
Change: How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH H2O --> Ca(OH) H2 Before moles excess moles moles Change moles - 5 moles moles moles ___________________________________________ After

19
After How many moles of hydrogen gas will be produced if 2.5 moles of calcium hydride react according to the following equation? CaH H2O --> Ca(OH) H2 Before moles excess moles moles Change moles - 5 moles moles moles ___________________________________________ After moles excess moles moles

20
Mole Ratios: #3 How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H O > CO H2O Before Change ____________________________________________________ After

21
Before How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H O > CO H2O Before mol excess mol 0 mol Change ____________________________________________________ After

22
Change How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H O > CO H2O Before mol excess mol 0 mol Change mol mol mol mol ____________________________________________________ After

23
After How many moles of water will be produced if 0.45 moles of oxygen gas reacts according to the following equation? 2 C6H O > CO H2O Before mol excess mol 0 mol Change mol mol mol mol ____________________________________________________ After mol excess mol mol

24
**Only moles go in the BCA table**

If given mass of reactants for products, convert to moles first, then use the table.

25
**Limiting reactant problems**

BCA approach distinguishes between what you start with and what reacts. When 0.50 mole of aluminum reacts with 0.72 mole of iodine The balanced equation deals with how many, not how much. to form aluminum iodide, how many moles of the excess reactant will remain? How many moles of aluminum iodide will be formed? 2 Al I2 ----> 2 AlI3 Before: Change After

26
**Limiting reactant problems**

Guess which reactant is used up first, then check 2 Al I2 ----> 2 AlI3 Before: Change After It’s clear to students that there’s not enough I2 to react with all the Al.

27
**Limiting reactant problems**

Now that you have determined the limiting reactant, complete the table, then solve for the desired answer. 2 Al I2 ----> 2 AlI3 Before: Change After

28
BCA a versatile tool It doesn’t matter what are the units of the initial quantities Mass - use molar mass Gas volume - use molar volume Solution volume - use molarity Convert to moles, then use the BCA table Solve for how many, then for how much

Similar presentations

Presentation is loading. Please wait....

OK

Stoichiometry with Chemical Reactions

Stoichiometry with Chemical Reactions

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on modern science and technology Ppt on power system stability classification Respiratory system anatomy and physiology ppt on cells Multifunction display ppt online Ppt on hard gelatin capsules Ppt on forward contract examples Ppt on duty roster boy Converter pub to ppt online student Ppt on carpooling system Ppt on wireless power supply