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Stoichiometry

Stoichiometry. The MOLE RATIO. In a balanced chemical reaction, the coefficients tell you how many moles of each substance you need for the reaction and how many moles of each product you will produce (theoretically). C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O

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Stoichiometry

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  1. Stoichiometry

  2. The MOLE RATIO • In a balanced chemical reaction, the coefficients tell you how many moles of each substance you need for the reaction and how many moles of each product you will produce (theoretically). • C3H8 + 5O2 3CO2 + 4H2O • 1mol C3H8 + 5mol O2  3mol CO2 + 4mol H2O

  3. What is the MOLE RATIO used for? • The mole ratio is used to convert from one substance to another. (Dimensional Analysis style of course!) We use this because we don’t have any instruments that measure things in MOLES!

  4. Example • How many moles of CO2 can theoretically be produced from 11.0 mol O2? • C3H8+ 5O2 3CO2 + 4H2O • 11.0 mol O2 3mol CO2 = 6.60 mol CO2 5mol O2

  5. Pause now to try some examples with your favorite chemistry teacher!

  6. Of course, we don’t measure things in moles and most of the time our answers aren’t in moles either. • We measure substances in… • Mass (grams) • Volume (mL) • Molecules/Atoms • Lucky for us, we know how to convert all of those into and out of moles!

  7. 2HCl + CaCO3 CO2 + H2O + 2CaCl2 • How many moles of calcium chloride are produced from 100.0g of blackboard chalk, calcium carbonate? • What are we given in the problem? • Balanced equation • 100.0g of CaCO3 • What are we trying to find? • Moles of CaCl2 • What should we do 1st?

  8. 2HCl + CaCO3 CO2 + H2O + 2CaCl2 • How many moles of calcium chloride are produced from 100.0g of blackboard chalk, calcium carbonate? • 1st – Convert your given into moles! • 200.0g CaCO3 1 mol CaCO3 = 2.000mol 100.g CaCO3 CaCO3

  9. 2HCl + CaCO3 CO2 + H2O + 2CaCl2 • How many moles of calcium chloride are produced from 200.0g of blackboard chalk, calcium carbonate? • Then, take that answer and use it with your mole ratio! • 2.000mol CaCO3 2mol CaCl2 = 4.000mol 1 mol CaCO3 CaCl2

  10. 2FeCl3 + 3H2S  Fe2S3 + 6HCl • What is the theoretical yield hydrochloric acid (in grams) produced when 8.53 × 1024 molecules of hydrosulfuric acid react with excess iron (III) chloride? • What are we given in the problem? • Balanced equation • 8.53 × 1024molecules H2S • What are we trying to find? • Mass of HCl produced

  11. 2FeCl3 + 3H2S  Fe2S3 + 6HCl • What is the theoretical yield hydrochloric acid (in grams) produced when 8.53 × 1024 molecules of hydrosulfuric acid react with excess iron (III) chloride? • 1st – Convert those molecules to moles! 8.53 × 1024molecules H2S 1 mole H2S = 14.2 mol H2S 6.02 × 1023 molecules H2S

  12. 2FeCl3 + 3H2S  Fe2S3 + 6HCl • What is the theoretical yield hydrochloric acid (in grams) produced when 8.53 × 1024 molecules of hydrosulfuric acid react with excess iron (III) chloride? • Now we can use the mole ratio! • 14.2 mol H2S 6 molHCl = 28.4 molHCl 3 mol H2S Finally, we can turn those moles into mass! 28.4 molHCl 36.5 g HCl = 1036.6g HCl = 1040g HCl 1 molHCl

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