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Page 1: Law of Conservation of Mass Lavoisier (1743-1794) 100 g Coal is completely burned What do you observe? STARTEND

Law of Conservation of MassLavoisier (1743-1794)

100 g 100 gCoal is completely

burned

What do you observe?

START END

Page 2: Law of Conservation of Mass Lavoisier (1743-1794) 100 g Coal is completely burned What do you observe? STARTEND

Law of Conservation of MassLavoisier (1743-1794)

In a chemical reaction, matter is neither created nor destroyed, it is transformed into something

else.

100 g 100 gCoal is completely

burnedWeight of jar & its contents remains the same. Mass is conserved

Page 3: Law of Conservation of Mass Lavoisier (1743-1794) 100 g Coal is completely burned What do you observe? STARTEND

Since matter can not be created or destroyed, chemical reactions must be __________ in terms of _________.

Reactants Products

The amount of mass you start with must

equal to the mass of the products:

100g total =

100g total

balanced mass

Page 4: Law of Conservation of Mass Lavoisier (1743-1794) 100 g Coal is completely burned What do you observe? STARTEND

According to the law of conservation of mass, how much zinc was present in the zinc carbonate?

A 40 g B 88 g C 104 g D 256 gREMEMBER !!!

The mass on both sides of the arrow must be equal.

+ =

256g 256g

+

ANSWER: 256- 152 = 104g of Zn

Page 5: Law of Conservation of Mass Lavoisier (1743-1794) 100 g Coal is completely burned What do you observe? STARTEND

CuCO3(s) CuO(s) + CO2(g)123.6 g 79.6 g ? g44.0 g

What are the reactants? CuCO3

What are the products?

CuO & CO2

What is the mass of reactants? 123.6 g

79.6 g + CO2

What is the mass of products?

Mass?

Answer: 123.6 – 79.6 = 44.0g of CO2

According to the law of conservation of mass, how much carbon dioxide was present in the copper (II) carbonate?

123.6g

+

123.6g

Page 6: Law of Conservation of Mass Lavoisier (1743-1794) 100 g Coal is completely burned What do you observe? STARTEND

For the Last Two Examples• You can set every Chemical Equation up like an

Algebraic Equation and solve it for the variable. For instance this equation,

can be written algebraically using the masses given on the previous slide like below. Now you only have to solve for x!

123.6 [g] = 79.6 [g] + x

CuCO3(s) CuO(s) + CO2(g)


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