To determine who is correct, let's analyze the equation for the reaction:
\[ 2H_2O \rightarrow 2H_2 + O_2 \]
This equation shows that 2 moles of water (H₂O) produce 2 moles of hydrogen gas (H₂) and 1 mole of oxygen gas (O₂).
First, we need to calculate the molar masses of the substances involved:
- Molar mass of H₂O = 2 (H) + 16 (O) = 18 g/mol
- Molar mass of H₂ = 2 (H) = 2 g/mol
- Molar mass of O₂ = 2 (O) = 32 g/mol
Now let's find the number of moles of hydrogen and oxygen collected in the lab:
- Moles of Hydrogen (H₂) collected:
\[ \text{Moles of } H_2 = \frac{112 \text{ g}}{2 \text{ g/mol}} = 56 \text{ mol} \]
- Moles of Oxygen (O₂) collected:
\[ \text{Moles of } O_2 = \frac{888 \text{ g}}{32 \text{ g/mol}} = 27.75 \text{ mol} \]
According to the stoichiometry of the reaction, for every 2 moles of H₂ produced, 1 mole of O₂ is produced. Therefore, the mole ratio is:
\[ \frac{H_2}{O_2} = \frac{2}{1} \]
Let's determine the number of moles of O₂ that would be expected from 56 moles of H₂:
\[ \text{Expected moles of } O_2 = \frac{56 \text{ mol H}_2}{2} = 28 \text{ mol O}_2 \]
We actually collected 27.75 moles of O₂, which is very close to the expected value. Thus, based on the amounts of gases collected, the reaction’s proportions are consistent.
Now, let's calculate the total mass of water that would have produced these volumes of hydrogen and oxygen gases. Using the stoichiometry:
- Each 2 moles of H₂O yield 2 moles of H₂ and 1 mole of O₂.
- Therefore, for 28 moles of O₂ produced, we would need:
\[ \text{Moles of } H_2O = 2 \times \text{moles of } O_2 = 2 \times 28 \text{ mol} = 56 \text{ mol of } H_2O \]
Now, we can find the mass of water:
\[ \text{Mass of } H_2O = 56 \text{ mol} \times 18 \text{ g/mol} = 1008 \text{ g} \]
In summary, the initial amount of water must have been roughly 1000 g, specifically about 1008 g.
Therefore, Student B is correct that the water must have been approximately 1000 g.