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[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Bellwork NEVER change subscripts
[object Object]
[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
N 2 (g)  +  3H 2 (g)     2NH 3 (g)  2 N  +  6 H  =  2 N  +  6H Atoms are conserved!!
Coefficients tell us the RATIO of molecules  N 2 (g)  +  3H 2 (g)     2NH 3 (g)
Coefficients tell us the RATIO of molecules  one N 2   +  three H 2      two NH 3 molecule  molecules  molecules   N 2 (g)  +  3H 2 (g)     2NH 3 (g)
Coefficients tell us the RATIO of moles  N 2 (g)  +  3H 2 (g)     2NH 3 (g)
one mole +  three moles    two moles N 2   H 2   NH 3   Coefficients tell us the RATIO of moles  N 2 (g)  +  3H 2 (g)     2NH 3 (g)
Find the molar mass of each compound & multiply by the coefficient   N 2 (g)  +  3H 2 (g)     2NH 3 (g)
28g   +  3 x 2g     2 x 17g N 2 (g)  +  3H 2 (g)     2NH 3 (g)  Find the molar mass of each compound & multiply by the coefficient
28g   +  3 x 2g     2 x 17g Why do you multiply the molar mass by three? N 2 (g)  +  3H 2 (g)     2NH 3 (g)  Find the molar mass of each compound & multiply by the coefficient
28g   +  3 x 2g     2 x 17g 28g  +  6g     34g  N 2 (g)  +  3H 2 (g)     2NH 3 (g)  Find the molar mass of each compound & multiply by the coefficient
34g = 34g Mass is conserved! 28g   +  3 x 2g     2 x 17g 28g  +  6g     34g  N 2 (g)  +  3H 2 (g)     2NH 3 (g)  Find the molar mass of each compound & multiply by the coefficient
N 2 (g)  +  3H 2 (g)     2NH 3 (g)  At STP, one mole of  gas occupies 22.4 liters.
22.4L N 2  +  3  22.4L H 2    2  22.4L   NH 3 N 2 (g)  +  3H 2 (g)     2NH 3 (g)  At STP, one mole of  gas occupies 22.4 liters.
N 2 (g)  +  3H 2 (g)     2NH 3 (g)  At STP, one mole of  gas occupies 22.4 liters. 22.4L N 2  +  3  22.4L H 2    2  22.4L   NH 3 22.4L 22.4L 22.4L 22.4L 22.4L 22.4L
N 2 (g)  +  3H 2 (g)     2NH 3 (g)  At STP, one mole of  gas occupies 22.4 liters. 22.4L N 2  +  67.2L H 2     44.8L   NH 3 22.4L N 2  +  3  22.4L H 2    2  22.4L   NH 3 22.4L 22.4L 22.4L 22.4L 22.4L 22.4L
[object Object]
[object Object]
[object Object],[object Object]
[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],2 moles of H 2 ( g )
[object Object],[object Object]
[object Object],[object Object],2 moles of O 2 ( g )
Conceptual Problem 12.1
Conceptual Problem 12.1
Conceptual Problem 12.1
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Lecture 12.1- Interpreting Balanced Equations

  • 1.
  • 2.
  • 3.
  • 4.
  • 5.
  • 6. N 2 (g) + 3H 2 (g)  2NH 3 (g) 2 N + 6 H = 2 N + 6H Atoms are conserved!!
  • 7. Coefficients tell us the RATIO of molecules N 2 (g) + 3H 2 (g)  2NH 3 (g)
  • 8. Coefficients tell us the RATIO of molecules one N 2 + three H 2  two NH 3 molecule molecules molecules N 2 (g) + 3H 2 (g)  2NH 3 (g)
  • 9. Coefficients tell us the RATIO of moles N 2 (g) + 3H 2 (g)  2NH 3 (g)
  • 10. one mole + three moles  two moles N 2 H 2 NH 3 Coefficients tell us the RATIO of moles N 2 (g) + 3H 2 (g)  2NH 3 (g)
  • 11. Find the molar mass of each compound & multiply by the coefficient N 2 (g) + 3H 2 (g)  2NH 3 (g)
  • 12. 28g + 3 x 2g  2 x 17g N 2 (g) + 3H 2 (g)  2NH 3 (g) Find the molar mass of each compound & multiply by the coefficient
  • 13. 28g + 3 x 2g  2 x 17g Why do you multiply the molar mass by three? N 2 (g) + 3H 2 (g)  2NH 3 (g) Find the molar mass of each compound & multiply by the coefficient
  • 14. 28g + 3 x 2g  2 x 17g 28g + 6g  34g N 2 (g) + 3H 2 (g)  2NH 3 (g) Find the molar mass of each compound & multiply by the coefficient
  • 15. 34g = 34g Mass is conserved! 28g + 3 x 2g  2 x 17g 28g + 6g  34g N 2 (g) + 3H 2 (g)  2NH 3 (g) Find the molar mass of each compound & multiply by the coefficient
  • 16. N 2 (g) + 3H 2 (g)  2NH 3 (g) At STP, one mole of gas occupies 22.4 liters.
  • 17. 22.4L N 2 + 3  22.4L H 2  2  22.4L NH 3 N 2 (g) + 3H 2 (g)  2NH 3 (g) At STP, one mole of gas occupies 22.4 liters.
  • 18. N 2 (g) + 3H 2 (g)  2NH 3 (g) At STP, one mole of gas occupies 22.4 liters. 22.4L N 2 + 3  22.4L H 2  2  22.4L NH 3 22.4L 22.4L 22.4L 22.4L 22.4L 22.4L
  • 19. N 2 (g) + 3H 2 (g)  2NH 3 (g) At STP, one mole of gas occupies 22.4 liters. 22.4L N 2 + 67.2L H 2  44.8L NH 3 22.4L N 2 + 3  22.4L H 2  2  22.4L NH 3 22.4L 22.4L 22.4L 22.4L 22.4L 22.4L
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Editor's Notes

  1. A cookie recipe tells you the number of cookies that you can expect to make from the listed amounts of ingredients. Using Models How can you express a cookie recipe as a balanced equation?
  2. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  3. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  4. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  5. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  6. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  7. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  8. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  9. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  10. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  11. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  12. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  13. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  14. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?
  15. The balanced chemical equation for the formation of ammonia can be interpreted in several ways. Predicting How many molecules of NH 3 could be made from 5 molecules of N 2 and 15 molecules of H 2 ?