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Conversions to the Max!.  The study of quantitative relationships between the amounts of reactants used and the amount of products formed by a chemical.

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Presentation on theme: "Conversions to the Max!.  The study of quantitative relationships between the amounts of reactants used and the amount of products formed by a chemical."— Presentation transcript:

1 Conversions to the Max!

2  The study of quantitative relationships between the amounts of reactants used and the amount of products formed by a chemical reaction

3  What do you need to know How to find molar mass of compounds…check! How to balance an equation…check! How to do a basic conversion…check!

4  4 Fe + 3 O 2 --> 2 Fe 2 O 3 First, to understand, let’s find the massof the reactants

5  4 mole Fe56 g Fe x = 224 g Fe 11 mole Fe 3 mole O 2 32 g O 2 x= 96 g O 2 11 mole O 2

6  224 g + 96 g = 320 g

7  2 mole Fe 2 O 3 160 g Fe 2 O 3 x 11 mole Fe 2 O 3 = 320 g Fe 2 O 3

8  Therefore, the mass of the products equals the mass of the reactants

9  If you have a balanced chemical equation, you only need to know the mass of ONE of the reactants or products in order to determine the mass of ALL others in the equation.

10  Each of the elements/compounds in the equation can be made into a mole ratio  Example: 2 K + Br 2  2 KBr 2 mole K = 1 mole Br 2

11  What are the other mole ratios in this equation? 2 mole K = 2 mole KBr 1 mole Br 2 = 2 mole KBr All of these can make a conversion factor…

12  Using the same equation: 2 K + Br 2  2 KBr  How many moles of potassium bromide would form if you start with 3.2 moles Br 2 ?

13 3.2 moles Br 2 2 moles KBr x 11 mole Br 2 = 6.4 moles KBr

14  Using the same equation: 2 K + Br 2  2 KBr  How many grams of potassium bromide would form if you start with 2.5 moles K?

15 2.5 mole K 2 mole KBr 119 g KBr x x 1 2 mole K 1 mole KBr = 297.5 g KBr

16  Using the same equation: 2 K + Br 2  2 KBr  How many grams of potassium bromide would form if you start with 25.7 g of potassium?

17 25.7 g K1 mole K 2 mole KBr 119 g KBr x x x 1 39 g K 2 mole K 1 mole KBr = 78.4 g KBr

18  Using the same equation: 2 K + Br 2  2 KBr  How many grams of potassium bromide would form if you start with 11.9 g of bromine?

19 11.9 g Br 2 1 mole Br 2 2 mole KBr 99 g KBr x x x 1 160 g Br 2 1 mole Br 2 1 moleKBr = 14.73 g KBr


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