se 402 3 conservation of mass

8
5.3 – To explain the law of conversation of matter in a chemical change after conducting an experiment involving a neutralization reaction CONSERVATION OF MASS

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Page 1: SE 402 3  Conservation of Mass

5.3 – To explain the law of conversation of matter in a chemical change after conducting an experiment involving a neutralization reaction

CONSERVATION OF MASS

Page 2: SE 402 3  Conservation of Mass

Chemical equations consist of reactants and products

Matter can be neither created nor destroyed

Reactants Products

2 H2 + O2 2 H2O

Page 3: SE 402 3  Conservation of Mass

…the mass of the reactants is always equal to the mass

of the products

2H2 + O2 2H2OReactants Products

4g + 32g = 36g 36 g = 36 g

Page 4: SE 402 3  Conservation of Mass

mass of reactants = mass of products

In a chemical reaction below, how many grams of oxygen (O2) was used given the following information.

29g ? 88g 48g

29 g + O2 = 88 g + 4829 g + O2 = 136 gO2 = 107 grams

10 24 2 22 + 13 8 + 10 OC O COH H

Page 5: SE 402 3  Conservation of Mass

Water (H2O) can be broken down into hydrogen and oxygen.

During a reaction you notice that 9 grams of water will produce 8 grams of oxygen (O2) and a certain amount of hydrogen (H2). How much hydrogen is produced?

9 g = ? + 8 g H2 = 1 grams

2 H2O 2 H2 +O2

2 H2O 2 H2 + O2

Page 6: SE 402 3  Conservation of Mass

37 grams sulphuric acid (H2SO4) is neutralized with calcium hydroxide (Ca(OH)2).

During the reaction, calcium sulphate (CaSO4), a salt, and water are produced. If 14g of water & 51g of salt are produced, how much calcium hydroxide was initially used?

37 g + Ca(OH)2 = 51 g + 14 g

Ca(OH)2 = 28 grams

H2SO4+ Ca(OH)2 CaSO4 + 2H2O

37 g + Ca(OH)2 = 65 g

Page 7: SE 402 3  Conservation of Mass

o Conservation of matter states the mass of the reactants will equal the mass of the products.

MASS OF REACTANTS = MASS OF PRODUCTS

Page 8: SE 402 3  Conservation of Mass

Study guide, Module 3, page 36 Worksheet # 10