Final answer:
To form 8.0 grams of water, the stoichiometry of the reaction between hydrogen and oxygen shows that less than 1 gram of hydrogen gas would be needed, specifically about 0.897 grams.
Step-by-step explanation:
To determine how many grams of hydrogen gas would be needed to form 8.0 grams of water, we should first understand the stoichiometry of the chemical reaction between hydrogen and oxygen to form water. The reaction is:
2 H2 + O2 → 2 H2O
The molecular weight of hydrogen (H2) is about 2.02 g/mol, and the molecular weight of water (H2O) is about 18.02 g/mol. According to the balanced equation, 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. If we have 8.0 g of water, we can calculate the moles of water and then use the stoichiometry of the reaction to find out the moles and subsequently the mass of hydrogen needed.
First, calculate the moles of water:
8.0 g H2O × (1 mole H2O / 18.02 g H2O) = 0.444 moles of H2O
Since the ratio of hydrogen to water is 1:1 in moles, 0.444 moles of water would need 0.444 moles of hydrogen gas. Now, calculate the mass of hydrogen gas:
0.444 moles H2 × (2.02 g H2 / 1 mole H2) = 0.897 g of H2
Therefore, to form 8.0 grams of water, 0.897 grams of hydrogen gas would be needed, which is not an option listed. Hence, there might be a mistake in the provided options or the question. However, based on the stoichiometry, the correct mass of hydrogen required should be less than 1 gram.