Please HELP! Use the following Equations to answer the problem: CH3OH + O2 —> CO2 + H2O

a) Balance the Equation:_CH3OH + _O2 + _H2O

The answer to your question is below

Explanation:

1)

Balanced chemical reaction

              2CH₃OH  + 3O₂  ⇒    2 CO₂  +  4H₂O

          Reactant            Element         Product

                2                         C                    2

                8                         H                    8

                8                         O                    8        

Molar mass of CH₃OH = 2[12 + 16 + 4]

                                     = 2[32]

                                     = 64 g

Molar mass of O₂ = 3[16 x 2] = 96 g

Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67

Experimental proportion CH₃OH/O₂ = 60/48 = 1.25

Conclusion

The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion

2)

Balanced chemical reaction

                         S₈  +  12O₂  ⇒    8SO₃

             Reactant     Elements     Products

                    8                  S                8

                   24                 O              24

Molar mass of S₈ = 32 x 8 = 256 g

Molar mass of O₂ = 12 x 32 = 384 g

Theoretical proportion S₈ / O₂ = 256 / 384

                                                  = 0.67

Experimental proportion S₈ / O₂ = 40 / 35

                                                     = 1.14

Conclusion

The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.          

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