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GSCI 163

GSCI 163. Lecture 7. Review. Elements combine in chemical reactions to produce new substances with different chemical properties. Balancing chemical equations require:

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GSCI 163

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  1. GSCI 163 Lecture 7

  2. Review • Elements combine in chemical reactions to produce new substances with different chemical properties. • Balancing chemical equations require: • Keeping the same number of elements on both sides of the chemical equation by adjusting coefficients (never subscripts of the chemical formula)

  3. Activity • Determine the chemical formula (yourself) of each substance in the following chart • Find the appropriate chemical equation (your partner in the reaction) • Balance the chemical equation for each reaction

  4. Find your partner and write the chemical reactions • Copper + sulfur  Copper(II) sulfide • Calcium oxide + water  Calcium hydroxide • Hydrogen + nitrogen  • Chromium + oxygen  • Aluminum + bromide  • Sodium + iodine  • Hydrogen + chlorine 

  5. How much is produced? • In a chemical reaction, a certain mass of reagents combine to produce a certain mass of products. Consider the following reaction: 2Na + I2 2NaI Here two atoms of sodium react with one molecule of iodine to produce two molecules of sodium iodide.

  6. Problem How much sodium iodide is produced when 10 grams of sodium react with 20 gram of iodine? • Questions that need to be answered: • How many atoms are there in 10 grams of Na? • How many molecules are there in 20 g of I2?

  7. Start with atomic weight 1 u = weight of 1/12 of 12C 2Na + I2 2NaI • Atomic weight of Na = 23u, iodine = 127u. Formula weight is the sum of the weight of the elements of the compound Formula weight of iodine molecule = 254u Formula weight of sodium iodide = 150u 2 x 23 u + 254 u = 2 x 150u 300 u = 300 u

  8. Moles and Avogadro’s number 23 g of Na (atomic weight = 23u) correspond to 6.02 x 1023atoms of Na and 1 mole of Na 254 g of I2 (formula weight = 254u) correspond to 6.02 x 1023molecules of I2 and 1 mole of I2 150 g of NaI (formula weight = 150u) correspond to 6.02 x 1023molecules of NaI and 1 mole of NaI Avogadro’s Number

  9. Number of atoms How many atoms are in 10 g of sodium? How many molecules in 20 g of iodine? = 26.3 x 1022 atoms 6.02 x 1023 atoms 6.02 x 1023 molecules 10 g x 20 g x = 2.63 x 1023 atoms = 4.74 x 1022 molecules 23 g 254 g

  10. Reaction 2Na + I2 2NaI Each molecule of iodine reacts with 2 atoms of sodium to produce 2 molecules of NaI Thus, 4.74 x 1022 molecules will react with 9.48 x 1022 atoms of sodium, producing 9.48 x 1022 molecules of NaI

  11. Final product • 16.8 x 1022 atoms of Na are left out. We say that one of the reagents (iodine) was used up. (This corresponds to 6.4 g of Na) To answer our initial question: • (9.48 x 1022) x 150 g / 6.02 x 1023 = 23.6 g of sodium iodide was produced

  12. Can we make this simpler? Use moles instead: 1 mole = 6.02 x 1023 atoms, molecules, etc Thus, • 23 g of Na is 1 mole  10 g of Na is 0.435 mole • 254 g of I2 is 1 mole  20 gof I2 is 0.079 mole

  13. Reaction 2Na + I2 2NaI Each molecule of iodine reacts with 2 atoms of sodium to produce 2 molecules of NaI Thus, 0.079 mole of iodine will react with 0.157 (2 x 0.079) mole of sodium, producing 0.157 mole of NaI

  14. Final product • 0.277 mole of Na are left out (0.435 - 0.157). We say that one of the reagents (iodine) was used up. (This corresponds to 6.4 g of Na) To answer our initial question: • 0.157 x 150 g = 23.6 g of sodium iodide was produced

  15. Recipe • Find the formula weights of the compounds involved in the reaction • Calculate the number of moles given by the actual mass of reagents • Calculate the number of moles of reagents used in the reaction • Calculate the number of moles of product produced • Convert the number of moles of the product to mass using the formula weight of the product Exercise: calculate the mass of water produced by the reaction of 10 g of hydrogen gas with 10 g of oxygen gas.

  16. How we classify reactions • Combination or synthesis X + Y  XY • Decomposition reaction XY  X + Y • Replacement reaction XY + Z  XZ + Y • Ion exchange reaction 3Ca(OH)2 + Al2(SO4)3 3CaSO4 + 2Al(OH)3 • Redox reaction – Gain and losses of electrons

  17. Types of chemical reaction

  18. Next class • Solutions and numerical representation.

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