If 2 grams of H₂(g) and 64 grams of O₂(g) were in a sealed container and the following reaction occurred: 2H₂(g) + O₂(g) ⇄ 2H₂O(g)...

For the given equation:

`2H_2 + O_2 harr2H_2O`

`H_2 + 1/2 O_2 harr H_2O`

We can use stoichiometry to determine the relationship between products and reactants. Here, 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water. In the second equation, 1 mole of hydrogen reacts with 0.5 mole of oxygen to produce 1 mole of water.

The molecular mass of hydrogen = 2 x 1 = 2 gm/mole

molecular mass of oxygen = 2 x 16 = 32 gm/mole and that of water = 2x1 + 16 = 18 gm/mole.

We are given 2 gm of hydrogen, that is 1 mole of hydrogen, and 64 gm of oxygen, that is 2 moles of oxygen. From the stoichiometry, we can see that 1 mole of hydrogen will react with only 0.5 mole of oxygen and will produce 1 mole of water.

That is, with the given quantity of reactants, only 1 mole or 18 gm of water will be produced.

Hope this helps.