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Tom [10]
4 years ago
13

Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these are decomposed, the sulfur dioxide produce

s 3.52 g oxygen and 3.53 g sulfur, while the sulfur trioxide produces 9.00 g oxygen and 6.00 g sulfur.
Calculate the mass of oxygen per gram of sulfur for each sample and show that these results are consistent with the law of multiple proportions
Chemistry
1 answer:
dexar [7]4 years ago
7 0

Answer:

See explanation

Explanation:

For sulphur dioxide;

3.53 g of sulphur combines with 3.52 g of oxygen

1.00 g of sulphur combines with 1.00 × 3.52/ 3.53 = 0.997 g

For sulphur trioxide

6.00 of sulphur combines with 9.00 g of oxygen

1.00 g of sulphur combines with 1.00 × 9.00/6.00 =1.5 g

Ratio of mass of oxygen;

1 : 1.5

The law of multiple proportions states that, if two elements A and B combine to form more than one chemical compounds, then the various masses of element A which combines separately with a fixed mass of element B are in simple multiple ratio.

We can see that the mass of oxygen which combines separately with sulphur in sulphur dioxide and sulphur trioxide are in simple ratio of 1:1.5.

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Answer:

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        

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                                     = 2[32]

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

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2)

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             Reactant     Elements     Products

                    8                  S                8

                   24                 O              24

Molar mass of S₈ = 32 x 8 = 256 g

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Theoretical proportion S₈ / O₂ = 256 / 384

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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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