The Arithmetic of Equations

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Presentation transcript:

The Arithmetic of Equations

Using Everyday Equations A balanced chemical equation provides the same kind of quantitative information that a recipe does. If we know that in order to manufacture a tricycle we need the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P), these would be the reactants needed to produce a tricycle, the product, then by balancing this equation we can figure out how many reactants we would need to manufacture any number of products.

Example Using F, S, W, H, and P and ending with the product FSW3HP2 (A tricycle): What would the equation be to produce one tricycle would be: F + S + 3W + H + 2P ----> FSW3HP2 This balanced equation is the “recipe” for a single tricycle.

Using Balanced Equations Chemists use balanced equations as a basis to calculate how much reactant is needed, or product is formed in a reaction. Stoichiometry – The calculations of quantities in chemical reactions. This allows chemists to tally the amounts of reactants and products using ratios of moles or representative particles.

Interpreting Chemical Equations A balanced equation can be interpreted in terms of different quantities, including number of atoms, molecules or moles; mass; and volume. The following equation shows the synthesis of ammonia (a common fertilizer) N2(g) + 3H2 (g) ---->2NH3(g) What are the total number of atoms, molecules, moles, mass and volume on both the reactant and product side of this equation? Reactant Side Product side Atoms 8 8 Molecules 4 2 Moles 4 2 Mass 34 g 34 g Volume 89.6 L 44.8 L

Mass Conservation in Chemical Reactions Mass and Atoms are conserved in every chemical reaction Let’s try one: Interpret the equation for the formation of water from its elements in terms of number of molecules and moles, and volume of gasses at STP. 2H2(g) + O2(g) ----> 2H2O (g) 2 molecules of H2 + 1 molecule of O2 ----> 2 molecules of H2O 2 moles of H2 + 1 mole of O2 ----> 2 moles of H2O 44.8 L of H2 + 22.4 L of O2 ----> 44.8 L of H2O

12.1 pg. 389 #'s 9-10 Read 12.2 pgs. 390-398