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Balancing Chemical Equations

Balancing Chemical Equations. Obeying the Law of Conservation of Matter/Mass. BCl 3 + P 4 + H 2  BP + HCl. Make a T-chart with reactants on the left & products on the right Tally the number of each atom of each element: R_ P B 1 1 Cl 3 1 P 4 1 H 2 1.

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Balancing Chemical Equations

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  1. Balancing Chemical Equations Obeying the Law of Conservation of Matter/Mass

  2. BCl3 + P4 + H2 BP + HCl • Make a T-chart with reactants on the left & products on the right • Tally the number of each atom of each element: R_ P B 1 1 Cl 3 1 P 4 1 H 2 1

  3. BCl3 + P4 + H2 BP + HCl • Notice on the T-chart a noticeable difference between Reactant & Product and start there • I would choose P. Write the coefficient 4 in front of BP so that the P’s balance. BCl3 + P4 + H2 4BP + HCl • Recalculate the T-chart. Note that now R’s Boron has 1, but P’s Boron now has 4. Place a coefficient of 4 in front of BCl3 4BCl3 + P4 + H2 4BP + HCl

  4. #1] 4BCl3 + P4 + H2 4BP + HCl • Update T-chart. R_ P B 4 4 Cl 12 1  12 P 4 4 H 2 1  12 Note Cl’s lack of balance. Place a 12 before HCl. 4BCl3 + P4 + H2 4BP + 12HCl Note new imbalance of H now. Fix this by adding 6 before H2. 4BCl3 + P4 + 6H2 4BP + 12HCl DONE!!!

  5. C2H2Cl4 + Ca(OH)2 C2HCl3 + CaCl2 + H2O • Fix T-chart. C H Cl Ca O R: 2 4 4 1 2 P: 2 3 5 1 1 Notice that R’s Cl is 4 & therefore P’s Clmust be a multiple of 4. The Cl in C2HCl3 is an odd #. Begin by adding a coefficient of 2 to create an even number of Cl. C2H2Cl4 + Ca(OH)2 2C2HCl3 + CaCl2 + H2O Recalculate the T-chart.

  6. #2] C2H2Cl4 + Ca(OH)2 2C2HCl3 + CaCl2 + H2O C H Cl Ca O R: 2 4 4 1 2 P: 4 4 8 1 1 Right away you can see that to balance C & Cl, we need to write a coefficient of 2 before the 1st cpd. 2C2H2Cl4 + Ca(OH)2 2C2HCl3 + CaCl2 + H2O C H Cl Ca O R: 4 6 8 1 2 P: 4 4 6 8 1 1 2 To fix O, add 2 before H2O. DONE!

  7. #3] (NH4)2Cr2O7 N2 + Cr2O3 + H2O • Fix T-chart. • N H Cr O R: 2 8 2 7 P: 2 2 2 4 Quick fix H 1st. (NH4)2Cr2O7 N2 + Cr2O3 + 4H2O N H Cr O R: 2 8 2 7 P: 2 8 2 7 DONE!

  8. #4] Zn3Sb2 + H2O  Zn(OH)2 + SbH3 Zn Sb H O R: 3 2 2 1 P: 1 1 5 2 You can start with Zn or Sb or both simultaneously. Zn3Sb2 + H2O  3Zn(OH)2 + 2SbH3 Zn Sb H O R: 3 2 2 1 P: 3 2 12 6

  9. Zn3Sb2 + H2O  3Zn(OH)2 + 2SbH3 • Notice that the ratio of H:O is still 2:1 in the new products. Adjust H2O’s coefficient and you’re done. • Zn3Sb2 + 6 H2O  3 Zn(OH)2 + 2 SbH3 • See? This really isn’t so hard after all. Take your time & switch between products’ and reactants’ total # of elements.

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