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kkurt [141]
3 years ago
7

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

90g H and 4.71g of O, is the substance water?​
Chemistry
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

                    No the substance is not water.

Explanation:

                   The balance chemical equation for the decomposition of water is as follow;

                                           2 H₂O = 2 H₂ + O₂

Step 1: <u>Calculate moles of H₂O;</u>

               Moles  =  Mass / M.Mass

               Moles  =  5.0 g / 18.01 g/mol

               Moles  =  0.277 moles of H₂O

Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

                        2 moles of H₂O produced  =  1 mole of O₂

So,

                  0.277 moles of H₂O will produce  =  X moles of O₂

Solving for X,

                     X =  0.277 mol × 1 mol / 2 mol

                     X =  0.138 moles of O₂

Also,

According to equation,

                        2 moles of H₂O produced  =  2 mole of H₂

So,

                  0.277 moles of H₂O will produce  =  X moles of H₂

Solving for X,

                     X =  0.277 mol × 2 mol / 2 mol

                     X =  0.227 moles of H₂

Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.138 mol × 31.99 g/mol

                 Mass  =  4.44 g of O₂

For H₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.227 mol × 2.01 g/mol

                 Mass  =  0.559 g of H₂

Conclusion:

                   From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

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Answer:

2C₂H₆ +  [7]O₂     →      [4]CO₂ + [6]H₂O

Explanation:

Chemical equation:

C₂H₆ +  O₂     →      CO₂ + H₂O

Balanced chemical equation:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Step 1:

2C₂H₆ +  O₂     →      CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C = 1

H = 12                                    H = 2

O = 2                                     O = 3

Step 2:

2C₂H₆ +  O₂     →      4CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 2

O = 2                                     O = 9

Step 3:

2C₂H₆ +  O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 2                                     O = 14

Step 4:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 14                                     O = 14

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