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Solving

Solving. Chemical Equation Word Problems. The Alphabet, In Alphabetical Order. Aitch Are Ay Bee Cue Dee Double U Ee Ef El Em En Ess. Ex Eye Gee Jay Kay Oh Pea See Tee Vee Wy Yu Zee. Sidney Harris. The Mole Concept. Review - The Mole.

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Solving

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  1. Solving Chemical Equation Word Problems

  2. The Alphabet, In Alphabetical Order Aitch Are Ay Bee Cue Dee Double U Ee Ef El Em En Ess Ex Eye Gee Jay Kay Oh Pea See Tee Vee Wy Yu Zee Sidney Harris

  3. The Mole Concept Review - The Mole Avogadro’s Number = 6.022 x 1023

  4. Counting Atoms • Chemistry is a quantitative science - we need a "counting unit.“ • The MOLE • 1 mole is the amount of substance that contains as many particles (atoms or molecules) as there are in 12.0 g of C-12.

  5. The Mole is Developed Avogadro’s number

  6. Amadeo Avogadro(1776 – 1856) Particles in a Mole 1 mole = 602213673600000000000000 or 6.022 x 1023 Amadeo Avogadro:(1766-1856) never knew his own number; it was named in his honor by a French scientist in 1909. its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895. ? quadrillions trillions billions millions thousands There is Avogadro's number of particles in a mole of any substance.

  7. Careers in Chemistry - Philosopher Q: How much is a mole? A: A mole is a quantity used by chemists to count atoms and molecules. A mole of something is equal to 6.02 x 1023 “somethings.” 1 mole = 602 200 000 000 000 000 000 000 Q: Can you give me an example to put that number in perspective? A: A computer that can count 10,000,000 atoms per second would take 2,000,000,000 years to count 1 mole of a substance.

  8. Counting to 1 Mole Is that right? A computer counting 10 million atoms every second would need to count for 2 billion years to count just a single mole. Lets look at the mathematics. 365 days 24 hours 60 min 60 sec x sec = 1 year = 31,536,000 sec 1 year 1 day 1 hour 1 min Therefore 1 year has 31,536,000 seconds or 3.1536 x 107 sec. A computer counting 10,000,000 atoms every second could count 3.153 x 1014 atoms every year. Finally, 6.02 x 1023 atoms divided by 3.1536 x 1014 atoms every year equals 1,908,929,477 years or approximately 2 billion years!

  9. How Big is a Mole? One mole of marbles would cover the entire Earth (oceans included) for a depth of three miles. One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills.

  10. Avogadro’s Number • A MOLE of any substance contains as many elementary units (atoms and molecules) as the number of atoms in 12 g of the isotope of carbon-12. • This number is called AVOGADRO’s number NA = 6.02 x 1023 particles/mol • The mass of one mole of a substance is called MOLAR MASS symbolized by MM • Units of MM are g/mol • Examples H2 hydrogen 2.02 g/mol He helium 4.0 g/mol N2 nitrogen 28.0 g/mol O2 oxygen 32.0 g/mol CO2 carbon dioxide 44.0 g/mol

  11. Using the Mole Concept to solve chemical equation word problems.

  12. Solving Chemical Equation Word Problems • A six step process that can be used to solve just about any • chemical equation word problem that involves finding out • how much of one substance is needed to react with another • substance, or how much of a product is the result of • a chemical reaction. • This process requires the following six steps (skills): • 1. writing chemical formulas (we have done) • 2. finding molar masses (we have done) • 3. balancing chemical equations (we have done) • 4. using coefficients to find molar equivalents (new) • (you have done) 5. setting up and using simple math ratio & proportion • (you have done) 6. cross multiplying and dividing

  13. Solving Chemical Equation Word Problems Use the (six) step by step process to solve the following problem: How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?

  14. Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Writing chemical formulas. Step 1: Identify the reactants and products and then write the chemical equation using the appropriate chemical symbols and formulas. Identify the reactants: nitrogen trihydride oxygen NH3 + O2 → Identify the products: nitrogen gas water N2 + H2O Write the chemical equation using appropriate chemical symbols and formulas. NH3 + O2 → N2 + H2O

  15. Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Finding molar masses. Step 2: Before balancing the equation, use your periodic table to find the mass of one mole of each element or compound. Write the molar masses ABOVE the element or compound. One mole = 17 g 32g 28 g 18 g (molar masses) NH3 + O2 → N2 + H2O

  16. Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Balancing chemical equations. Step 3: Balance the equation using correct coefficients. One mole =17 g 32g 28 g 18 g(molar masses) 4 NH3 + 3 O2 → 2 N2 + 6 H2O

  17. Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Using coefficients to find molar equivalents. Step 4: Multiply the molar masses by the coefficients to get the balanced mass equivalents. Write this number below the symbol of the element or compound. One mole =17 g 32g 28 g 18 g(molar masses) 4 NH3 + 3 O2 → 2 N2 + 6 H2O 68 g 96 g 56 g108 g

  18. Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Setting up and using simple math ratio & proportion (fractions). Step 5: Using your problem statement and the balanced mass equivalents, set up a fractional ratio and proportion to solve for your unknown. A. Identify the unknown (the quantity you are trying to find = How much nitrogen gas in grams?) and the information that you do know (you have 10 grams of ammonia) and line them up right under the balanced mass equivalents for the chemicals that represent the same measurements. 4 NH3 + 3 O2 → 2 N2 + 6 H2O 68 g 96 g 56 g108 g 10 g X →

  19. B. Since we only need two fractions to solve for one unknown, we can now set up a simple equation to solve. (NOTE: Since oxygen and water are not part of this particular problem, we need them to balance the equation, but do not need them for the final solution.) 68 g= 56 g 10 g X

  20. Problem:How much nitrogen gas (grams) is released when ten grams of ammonia (nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water? Skill: Cross multiplying and dividing. Step 6: Cross multiply and divide to solve for the unknown (X) quantity. 68 g= 56 g 10 g X a. 68 g x X = 10 g x 56 g b. 68 X g = 560 g g c. 68 X g = 560 g g X = 8.2 g N2 68 g 68 g

  21. Solving Chemical Equation Word Problems Practice problem # 1: Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to make water. If you start with 100 grams of hydrogen gas and all the oxygen that you need to completely combine with the hydrogen, then how much water will be produced? hydrogen gas + oxygen gas → water 1 mol = 2 g 32 g 18 g H2 + O2→ H2O 2 2 4 g + = 36 g 100 g X X = 900 g

  22. Solving Chemical Equation Word Problems Practice problem # 2: You have 100 grams of sodium that you are going to combine with chlorine gas (diatomic) to make sodium chloride (salt). How much chlorine gas will you need to just exactly combine with the 100 grams of sodium? sodium + chlorine gas → sodium chloride 1 mol = 23 g 70 g 58 g Na + Cl2→ NaCl 2 2 46 g = 70 g 100 g X X = 152 g

  23. Your turn! Complete and turn in the worksheet (Due next class).

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