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Quantitative Information from Balanced Chemical Equations

Quantitative Information from Balanced Chemical Equations. Equation: 2H 2 (g) + O 2 (g) 2H 2 O(l ) The coefficients in the equation can mean the following. 2 molecules H 2 + 1 molecule O 2  2 molecules H 2 O

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Quantitative Information from Balanced Chemical Equations

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  1. Quantitative Information from Balanced Chemical Equations Equation: 2H2(g) + O2 (g) 2H2O(l) The coefficients in the equation can mean the following. • 2 molecules H2 + 1 molecule O2 2 molecules H2O • 2 moles H2 + 1 mole O2 2 moles H2O • Coefficients are not equivalent to grams! • The coefficients in the balanced equation give the ratio of moles of reactants and products 1 mol O2 2 mol H2 2 mol H2 2 mol H2O 2 mol H2O 1 mol O2

  2. Practice: Mole-to-Mole Relationships in Stoichiometric Calculations Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg(s) + N2(g)  Mg3N2 (s) The reaction of 3.5 moles of Mg will produce how many moles of Mg3N2? The reaction of 6.25 mole of Mg will require how many moles of N2?

  3. Stoichiometric Calculations Going to or from grams From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant). Figure 3.16 Procedure for calculating amounts of reactants consumed or products formed in a reaction.

  4. 1 mol C8H18 1 mol C8H18 16 mol CO2 25 mol O2 2 mol C8H18 2 mol C8H18 Example: Predicting Amounts from Stoichiometry The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance. How much CO2 in grams can be made from 22.0 g of C8H18 in the combustion of C8H18? How much O2 is used in this reaction? 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) 44.0 g CO2 22.0 g C8H18 1 mol CO2 114 g C8H18 = 67.9 g CO2 = 77.2 g O2 32.0 g O2 22.0 g C8H18 114 g C8H18 1 mol O2

  5. Practice: Stoichiometric Calculations Small amounts of O2 is prepared in the laboratory using KClO3. 2 KClO3(s) → 2 KCl(s) + 3 O2(g). How many grams of O2 can be prepared from 4.50 g of KClO3?

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