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Chapter 12: Chemical Quantities

Chapter 12: Chemical Quantities. Section 12.2: Using Moles. Balanced chemical equations. Used to relate moles of one substance to moles of another substance Use coefficients in the equation to convert to moles of other reactants or products. The following recipe serves 4:

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Chapter 12: Chemical Quantities

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  1. Chapter 12: Chemical Quantities Section 12.2: Using Moles

  2. Balanced chemical equations • Used to relate moles of one substance to moles of another substance • Use coefficients in the equation to convert to moles of other reactants or products

  3. The following recipe serves 4: 4 potatoes (.3 lb each) 2 onions (.2 lb each) 8 carrots (.1 lb each) 4 stalks of celery (.05 lb each) 1.4 lbs of water What is the total mass of the soup? How would you make enough for 8? How about 1240? **With a balanced chemical equation and number of moles, we can predict the exact amount of reactant and product in a reaction**

  4. There are 4 steps to follow: 1) Write the balanced chemical equation 2) Convert the given mass or volume to moles 3) Use the coefficients in the chemical equation to set up a mole ratio (the coefficients are the # of moles!) HINT: The substance you are solving for goes on TOP 4) Convert these moles back to mass or volume as required

  5. Steps Use: Grams A ↔ Moles A ↔ Moles B ↔ Grams B molar mass coefficients molar mass

  6. N2 + 3H2 2NH3 • This reaction can be stated as: 1 mole of nitrogen will react with 3 moles of hydrogen to produce 2 moles of ammonia.

  7. Practice Problems (pg. 415) #10) What mass of CO2 forms when 95.6 g of C3H8 burns? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) 3 C = 36.033 g C= 12.011g 8 H= 8.064 g 2 O = 32 g C3H8 = 44.097 g/mol CO2 = 44.011 g/mol 95.6 g C3H8 x 1 mol C3H8 x 3 mol CO2 x 44.011 g CO2= 44.097g C3H8 1 mol C3H8 1 mol CO2 = 286.2 g CO2

  8. Practice (cont) #11) How many grams of fluorine are required to produce 10 g XeF6? Xe (g) + 3F2 (g) → Xe F6 (s) F2: 2 x 18.998 g = 37.996 g Xe F6: 131.29 + (6 x 18.998) = 245.278 g 10 g XeF6 x 1 mol XeF6 x 3 mol F2 x 37.996 g F2 = 245.278 g XeF6 1 mol XeF6 1 mol F2 = 4.65 g F2

  9. LIMITING REACTANT The reactant that runs out first in a reaction/ stops the reaction

  10. Practice Problems #1) What is the limiting reactant in producing water (H2O), 5 g H2 or 5 g of O2? (convert g reactant → mol of product) 2H2 + O2 → 2H2O 5 g H2 x 1 mol H2 x 2 mol H2O = 2.48 mol H2O 2.0158 g H2 2 mol H2 5 g O2 x 1 mol O2 x 2 mol H2O = 0.31 mol H2O 31.998 g O2 1mol O2 Limiting reactant is O2 (runs out first)

  11. Practice (cont) 2) Which is the limiting reactant in producing NH3, 3.75 g N2 or 3.75 g H2? N2 + 3H2 → 2 NH3 3.75g H2 x 1 mol H2 x 2 mol NH3 = 1.24 mol NH3 2.0158 g H2 3 mol H2 3.75g N2 x 1 mol N2 x 2 mol NH3 = 0.27 mol NH3 28 g N2 1 mol N2 Limiting reactant is N2

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