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

c is the answer because when something is left after something evaporates that shows the chemical change.