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Question

Question
Under certain conditions, water can be separated into oxygen and hydrogen gases according to the equation: 2H2O → 2H2 + O2
.

In a lab, 96g of hydrogen gas and 404g of oxygen gas were collected. Student A says that the starting amount of water cannot be determined, but Student B says that the water must have been 500g. Who is correct?

(1 point)
Responses

Student A, because the water was not measured to start with.
Student A, because the water was not measured to start with.

Neither are correct. It was 2000g because the equation shows 2H2O
.
Neither are correct. It was 2000g because the equation shows 2 cap h sub 2 cap o.

Student B, because the law of conservation of matter.
Student B, because the law of conservation of matter.

Neither are correct. It would have been 1112g, because there are 2 H2
and 1 O2
.
Neither are correct. It would have been 1112g, because there are 2 H2
and 1 O2
.
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Answer 1

To determine who is correct between Student A and Student B regarding the amount of water that was present initially, we can use the law of conservation of mass and the balanced chemical equation provided.

The chemical equation is:

\[ 2H_2O → 2H_2 + O_2 \]

From the equation, we can see that 2 moles of water (H₂O) produce 2 moles of hydrogen (H₂) and 1 mole of oxygen (O₂).

Now, let's calculate the number of grams of hydrogen and oxygen produced:

1. Molar Masses:
   • Hydrogen (H) has a molar mass of about 1 g/mol, so H₂ (hydrogen gas) has a molar mass of about 2 g/mol.
   • Oxygen (O) has a molar mass of about 16 g/mol, so O₂ (oxygen gas) has a molar mass of about 32 g/mol.

Using this information:

• The total mass of the hydrogen collected is: \[ 96 \text{ g of } H_2 \]

• The total mass of the oxygen collected is: \[ 404 \text{ g of } O_2 \]

2. Calculate Moles:

   • Moles of hydrogen produced: \[ \text{Moles of } H_2 = \frac{96 \text{ g}}{2 \text{ g/mol}} = 48 \text{ moles} \]
   • Moles of oxygen produced: \[ \text{Moles of } O_2 = \frac{404 \text{ g}}{32 \text{ g/mol}} = 12.625 \text{ moles} \]

3. Relate Moles of Water to Hydrogen and Oxygen: From the balanced equation:

   • For every 2 moles of water, 2 moles of hydrogen and 1 mole of oxygen are produced: Therefore, if we have 12.625 moles of \( O_2 \), we need 25.25 moles of \( H_2O \) (because it takes 2 moles of water to produce 1 mole of \( O_2 \)). \[ \text{Moles of } H_2O = 2 \times \text{Moles of } O_2 = 2 \times 12.625 \approx 25.25 \text{ moles of } H_2O \]

4. Calculate Mass of Water: \[ \text{Mass of } H_2O = 25.25 \text{ moles} \times 18 \text{ g/mol} = 454.5 \text{ g} \]

Now looking at the original question, it seems that neither student is completely correct.

• Student A is incorrect, as the starting amount of water can indeed be calculated using the amounts of hydrogen and oxygen produced.
• Student B is also incorrect, as the calculated starting amount of water was 454.5 g, not 500 g.

Given this analysis, the best choice from the responses is that neither Student A nor Student B is correct. However, since there is no given option about 454.5 g, the closest incorrect option considering stoichiometry could be (which must not be strictly taken as accurate based on our calculation) that it would be 1112g, but that also does not reflect actual calculations.

In conclusion, both students are incorrect as neither provided the correct value for the mass of water.

Answer 2

what is the answer

Answer 3

The correct answer based on the calculations is:

Neither are correct. It would have been 1112g, because there are 2 H₂ and 1 O₂.

While the calculated amount of water was approximately 454.5 g, in the context of the provided answers and how they address the relationship in the balanced equation (2 H₂O → 2 H₂ + O₂), this option at least aligns somewhat with the stoichiometric ratios, even though it does not accurately reflect the actual results found from the experiment.

To summarize, based on the context of the options given, you would select:

Neither are correct. It would have been 1112g, because there are 2 H₂ and 1 O₂.