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Balancing Chemical Equations and Calculating Reactant Masses

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Learn how to balance chemical equations and calculate reactant masses step by step with examples and atomic mass calculations.

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Balancing Chemical Equations and Calculating Reactant Masses

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  1. First, write the equation.

  2. Count the number of atoms on each side of the equation. 1+2=3 Na - _____ H - _____ O - _____ 1 Na - _____ H - _____ O - _____ 1 2 3 1 1

  3. Count the number of atoms on each side of the equation. 2X2=4 2X1=2 2 Na - _____ H - _____ O - _____ 1 Na - _____ H - _____ O - _____ 1 = 2 4 3 1 2 1 =

  4. Count the number of atoms on each side of the equation. 2+2=4 2 2 2 Na - _____ H - _____ O - _____ 1 Na - _____ H - _____ O - _____ 1 = 2 4 3 4 1 2 1 2 =

  5. Count the number of atoms on each side of the equation. 2 2 2 Na - _____ H - _____ O - _____ 1 = Na - _____ H - _____ O - _____ 1 2 4 3 4 = 1 2 1 2 =

  6. Count the number of atoms on each side of the equation. 2 2 2 2 Na - _____ H - _____ O - _____ 1 2 = Na - _____ H - _____ O - _____ 1 2 4 3 4 = 1 2 1 2 =

  7. To find the mass of the reactants: • Find each element on the Periodic Chart • Find the Atomic Mass for each element • Add the atomic mass for each atom to get the total mass for the reactants

  8. 46 Na (sodium) has an atomic mass of 23. There are 2 Na (sodium) atoms. 2x23= 46

  9. 2x2=4 46 4 H (hydrogen) has an atomic mass of 1. There are 4 H (hydrogen) atoms. 4x1-4

  10. 46 4 +32 2x16=32 O (oxygen) has an atomic mass of 16. There are 2 O (oxygen) atoms. 2x16=32

  11. 46 4 +32 46 + 4 +32 = 82

  12. 82

  13. 46 Na (sodium) has an atomic mass of 23. There are 2 Na (sodium) atoms. 2x23= 46

  14. 46 +32 O (oxygen) has an atomic mass of 16. There are 2 Na (sodium) atoms. 2x16= 32

  15. 46 +32 +2 H (hydrogen) has an atomic mass of 1. There are 2 H (hydrogen) atoms. 2x1= 2

  16. 46 +32 +2 +2 H (hydrogen) has an atomic mass of 1. There are 2 H (hydrogen) atoms. 2x1= 2

  17. 46 +32 +2 +2 46+32+2+2=82

  18. 82 82 YES

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