Conversions to the Max!

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

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

 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

 224 g + 96 g = 320 g

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

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

 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.

 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

 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…

 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 ?

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

 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?

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

 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?

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

 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?

11.9 g Br 2 1 mole Br 2 2 mole KBr 99 g KBr x x x g Br 2 1 mole Br 2 1 moleKBr = g KBr