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April 3, 2014

April 3, 2014. Stoichiometry. Stoichiometry. Stoichiometry is the study of quantities of materials consumed and produced in chemical reactions Stoikheion (Greek, “element”) + - metry (English, measurement). Stoichiometry.

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April 3, 2014

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  1. April 3, 2014 Stoichiometry

  2. Stoichiometry Stoichiometry is the study of quantities of materials consumed and produced in chemical reactions Stoikheion (Greek, “element”) + -metry (English, measurement)

  3. Stoichiometry • Stoichiometry is used to make predictions about how much product will be formed without actually doing the reaction. • What you need: • A balanced equation!!

  4. 2 H2 + O2 2 H2O 2 molecules of H2 react with 1 molecule of O2 to form 2 molecules of water IT ALSO MEANS 2 moles of H2 react with 1 mole of O2 to form 2 moles of water

  5. Mole Ratios • Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H2+ O2 2 H2O • Hydrogen to oxygen: 2 mol H2 1 mol O2 • Hydrogen to water: • Oxygen to water: • Oxygen to hydrogen: • Water to oxygen:

  6. Mole Ratios • Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H2+ O2 2 H2O • Hydrogen to oxygen: 2 mol H2 1 mol O2 • Hydrogen to water: 2 mol H2 2 molH2O • Oxygen to water: • Oxygen to hydrogen: • Water to oxygen:

  7. Mole Ratios • Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H2 + O2 2 H2O • Hydrogen to oxygen: 2 mol H2 1 mol O2 • Hydrogen to water: 2 mol H2 2 molH2O • Oxygen to water: 1mol O2 2 mol H2O • Oxygen to hydrogen: • Water to oxygen:

  8. Mole Ratios • Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H2+ O2 2 H2O • Hydrogen to oxygen: 2 mol H2 1 mol O2 • Hydrogen to water: 2 mol H2 2 molH2O • Oxygen to water: 1mol O2 2 mol H2O • Oxygen to hydrogen: 1mol O2 2 mol H2 • Water to oxygen:

  9. Mole Ratios • Once you have a balanced reaction, you can use mole ratios to convert from amount of one substance to amount of another substance 2 H2 + O2 2 H2O • Hydrogen to oxygen: 2 mol H2 1 mol O2 • Hydrogen to water: 2 mol H2 2 molH2O • Oxygen to water: 1mol O2 2 mol H2O • Oxygen to hydrogen: 1mol O2 2 mol H2 • Water to oxygen: 2 mol H2O 1 mol O2

  10. Calculating Masses of Reactants and Products • Make sure the equation is balanced. • Identify your given and your unknown. • Convert mass to moles, if necessary. • Set up mole ratios. • Use mole ratios to calculate moles of the product/reactant desired. • Convert moles to mass, if necessary.

  11. Important • The “given” and “unknown” can both be reactants, both be products, or one reactant and one product • Reaction stoichiometry will ALWAYS have a mol (given) mol (unknown)conversion!!! • To calculate mol (given) mol (unknown), use mole ratio, which comes directly from the coefficients of the balanced chemical equation • Other conversion factors may (or may not) be necessary before or after this conversion

  12. MOL (GIVEN)  MOL (UNKNOWN) • 1. 2H2+ O22H2O • A. How many moles of water are produced from 1 mole of H2? Mole ratio H2 to H2O : 2 molH2=2 molH2O 1 molH2 | 2 molH2O = 1 x 2 = 1 mol H2O 2 molH2 2

  13. MOL (GIVEN)  MOL (UNKNOWN) • 1. 2H2+ O22H2O • B. How many moles of water are produced from 3 moles of O2? Mole ratio O2 to H2O : 1mol O2=2 mol H2O 3mol O2 | 2 mol H2O  = 3 x 2 = 6mol H2O 1 mol O2 1

  14. MOL (GIVEN)  MOL (UNKNOWN)  MASS (UNKNOWN) • 2. CH4+ 2O2 CO2 + 2H2O • A. How many grams of water are produced from one mole of CH4? (molar mass H2O = 18.02 g/mol) Mole ratio CH4 to H2O : 1 mol CH4= 2 mol H2O 1 mol CH4 | 2 mol H2O | 18.02 g = 1 x 2 x 18.02 = 36.04 g H2O 1 mol CH41 mol H2O 1 x 1

  15. MOL (GIVEN)  MOL (UNKNOWN)  MASS (UNKNOWN) • 2. CH4+ 2O2CO2 + 2H2O • B. How many grams of O2 are required to produce 5 mol CO2? (molar mass O2 = 32.00 g/mol) Mole ratio CO2 to O2 : 1 mol CO2= 2 mol O2 5mol CO2 | 2 mol O2| 32.00 g = 5 x 2 x 32.00 = 320 g O2 1 mol CO21 molO2 1 x 1

  16. mass, g (given) mol (given) mol (unknown) • 3. CH4+ 2O2CO2 + 2H2O • A. How many moles of CO2 can be produced from 4.0 g CH4? (molar mass CH4 = 16.05 g/mol) Mole ratio CH4 to CO2 : 1 mol CH4= 1mol CO2 4.0g CH4 | 1 molCH4| 1 mol CO2= 4.0 x 1 x 1 = 0.25mol CO2 16.05 g CH41 molCH416.05 x 1

  17. mass, g (given) mol (given) mol (unknown) • 3. CH4+ 2O2 CO2 + 2H2O • B. How many moles of CH4 are required to react with 2.0 g O2? (molar mass O2 = 32.00 g/mol) Mole ratio O2 to CH4: 2mol O2= 1 mol CH4 2.0 g O2 | 1 molO2  | 1 mol CH4= 2.0 x 1 x 1 = 0.031 mol CH4 32.00 g O22 mol O232.00 x 2

  18. mass, g (given) mol (given) mol (unknown)  mass, g (unknown) • 4. ____NH3 + ____O2____NO + ____H2O First you must balance the equation. N: 1 = N: 1 H: 3 ≠ H: 2 O: 2 = O: 2

  19. mass, g (given) mol (given) mol (unknown)  mass, g (unknown) • 4. __2__NH3 + ____O2____NO + __3__H2O N: 1 2 ≠ N: 1 H: 36 ≠ H: 2 6 O: 2 ≠ O: 3 4

  20. mass, g (given) mol (given) mol (unknown)  mass, g (unknown) • 4. __2__NH3 + ____O2_2__NO + __3__H2O N: 1 2 = N: 1 2 H: 36 = H: 2 6 O: 2 ≠ O: 34 5 Leave oxygen for last!

  21. mass, g (given) mol (given) mol (unknown)  mass, g (unknown) • 4. __4__NH3 + __5__O2_4__NO + __6__H2O N: 12 4 = N: 12 4 H: 36 12 = H: 26 12 O: 2 10 ≠ O: 34 5 10 There is no way to fit 2 evenly into 5. Try doubling everything and then see if you can make oxygen balance.

  22. mass, g (given) mol (given) mol (unknown)  mass, g (unknown) • 4. 4 NH3+ 5 O24 NO + 6 H2O • A. How many grams of H2O can be formed from 1.0 g NH3? (molar mass NH3 = 17.04 g/mol, molar mass H2O = 18.02 g/mol) Mole ratio NH3 to H2O: 4mol NH3= 6mol H2O 1.0g NH3| 1 molNH3 | 6 mol H2O | 18.02 g = 1.0 x 1 x 6 x 18.02 = 1.59 g H2O 17.04 g NH34 mol NH31 mol H2O 17.04 x 4

  23. mass, g (given) mol (given) mol (unknown)  mass, g (unknown) • 4. 4 NH3+ 5 O24 NO + 6 H2O • B. If 5.0 g NO are formed, how many grams of H2O are also formed? (molar mass NO = 30.01 g/mol) Mole ratio NO to H2O: 4 mol NO = 6 mol H2O 5.0 g NO| 1 mol NO | 6 mol H2O | 18.02 g = 5.0 x 1 x 6 x 18.02 = 4.50 g H2O 30.01 g NO4 mol NO1 mol H2O 30.01 x 4

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