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Chemisty chapter 10

Chemisty chapter 10. Background information. At STP volume of a gas is 22.4 dm 3 per mole Coefficients in a balanced equation tell the chemist three things about the quantities of reacctants and products A. the relative number of particles( atoms, molecules, formula units

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Chemisty chapter 10

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  1. Chemisty chapter 10

  2. Background information • At STP volume of a gas is 22.4 dm3 per mole • Coefficients in a balanced equation tell the chemist three things about the quantities of reacctants and products • A. the relative number of particles( atoms, molecules, formula units • B. The relative number of moles

  3. Background info • C. The relative volume of gases • Mole ratio the number of moles of substance given in problem and the number of moles of substance looking for in problem.

  4. Sample problem #2 • Carbon monoxide burns in oxygen to produce CO2. How many molecules of oxygen would react with 5.0 X 105 molecules of CO? • 5.0 X105 molecules CO x 1 molecule O2 2 molecules CO = 2.5 X 105 Molecules of O2

  5. Sample 3 • What volume of CO reacting with oxygen, would produce 200 dm3 of CO2 at constant temperature and pressure? • 200 dm3 CO2 X 2 dm3 CO = 200 dm3 CO • 2 dm3 CO2

  6. Mass problems solving • Step 1 Change the mass of the substance given in the statement of the problems to moles ( dividing the given mass of the substance by its molar mass) • Step 2 Change from moles of the substance given to moles of the substance sought. Use the coefficients in the balanced equation to do this.

  7. Mass problem solving • Step 3 Change from moles of the substance sought to mass of the substance sought. To do this multiple “ moles of the substance sought by the molar mass of the substance.

  8. Sample 4 • Determine the mass of NaCl that will decompose to yield 355 grams of Cl2. • 2NaCl - 2Na + Cl2 • 355 g Cl2 X 1 mole Cl2 X 2 moles NaCl X • 71 g of Cl2 1 mole Cl2 • 58 .5 g NaCl = 5.85 X 102 g NaCl • 1 mole NaCl

  9. Sample 5 • Calcium carbonate a solid reacts with dilute HCl to produce Carbon dioxide calcium chloride and water. What volume of CO2 measured at STP will be produced when 8.0 X 101 g of CaCO3 reacts?

  10. Sample 5 • CaCO3 + 2HCl  CO2 = CaCl2 + H2O • 8.0 X101g CaCO3 X 1 mole of CaCO3 X • 100 g of CaCO3 • 1 mole CO2 X 22.4 dm3 CO2 = 1 mole CaCO3 1 mole CO2 18 dm3 CO2

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