Final answer:
A 38.4 gram sample of hydrogen peroxide decomposing would yield approximately 20.34 grams of water, as determined from calculations based on molar masses and the stoichiometry of the decomposition reaction.
Step-by-step explanation:
The question deals with the decomposition of hydrogen peroxide (H2O2) into water (H2O) and oxygen (O2). The balanced chemical equation for this reaction is:
2H2O2 → 2H2O + O2
From the equation, we can see that two moles of hydrogen peroxide produce two moles of water. Since the molar mass of hydrogen peroxide is approximately 34.0147 g/mol, we can calculate the amount of water produced by first finding the number of moles of hydrogen peroxide in a 38.4 gram sample.
Number of moles of H2O2 = mass / molar mass = 38.4 g / 34.0147 g/mol ≈ 1.129 moles
Since the mole ratio of hydrogen peroxide to water is 1:1, 1.129 moles of hydrogen peroxide will produce 1.129 moles of water. The molar mass of water is approximately 18.01528 g/mol. Thus:
Mass of water = number of moles × molar mass = 1.129 moles × 18.01528 g/mol ≈ 20.34 grams
Therefore, if a 38.4 gram sample of hydrogen peroxide completely decomposed, approximately 20.34 grams of water would remain.