2H(2)+0(2)->2h(2)0

If I start with 5 H2 molecules and 5 O2 molecules, how many H and O atoms do I have?

How many molecules of water can I make?

Will I have any atoms leftover, and how much?

The answer to your question is below

Explanation:

Data

5 molecules of H₂

5 molecules of O₂

Balanced chemical reaction

                2H₂  +  O₂  ⇒   2H₂O

1.-

Atoms of H₂

                1 mol of H 6.023 x 10²³ atoms

                5 moles of H  x

                 x = (5 x 6.023 x 10²³) / 1

                 x = 3.012 x 10²⁴ atoms of H₂

Atoms of O₂

                1 mol of O 6.023 x 10²³ atoms

                5 moles of O  x

                x = (5 x 6.023 x 10²³) / 1

                x = 3.012 x 10²⁴ atoms of O₂

2.- Molecules of water

Limiting reactant  

Theoretical proportion H₂/O₂ = 2/1 = 2

Experimental proportion H₂/O₂ = 5/5 = 1

Conclusion

The limiting reactant is H₂ because the experimental proportion diminishes.

                    2 moles of H₂ 2 moles of H₂O

                    5 moles of H₂ x

                     x = (5 x 2)/2

                    x = 5 moles of Water

3.- Excess reactant

                    2 moles of H₂ 1 mol of O₂

                    5 moles of H₂ x

                    x = (5 x 1)/2

                    x = 2.5 moles of O₂ that reacted

Excess Oxygen = 5 moles - 2.5              

                          = 2.5 moles leftover

it should be all of them because it asks if you included all of those in your answer

explanation:

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error