How many grams of oxygen gas are required to produce 7.46 kJ of heat when hydrogen gas burns at constant pressure to produce gaseous water?

2 H2(g) + O2(g) → 2 H2O(g) ΔH = −484 kJ
Liquid water has a heat of vaporization of 44.0 kJ per mole at 25°C.

To determine the amount of oxygen gas required to produce 7.46 kJ of heat, we need to use the balanced equation and the enthalpy change of the reaction.

The balanced equation is: 2 H2(g) + O2(g) → 2 H2O(g)

The enthalpy change, ΔH, of the reaction is -484 kJ.

Since the balanced equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor, we can use stoichiometry to calculate the moles of oxygen gas required.

Step 1: Convert the heat change from kJ to J.
7.46 kJ = 7.46 * 1000 J
7.46 kJ = 7460 J

Step 2: Calculate the moles of water vapor produced.
Using the enthalpy change, we know that 484 kJ is released when 2 moles of water vapor are produced.
So, 7460 J (the heat produced) corresponds to (2/484) * 7460 moles of water vapor.
Moles of water vapor = (2/484) * 7460 = 30.99 moles

Step 3: As per the stoichiometry in the balanced equation, 1 mole of oxygen gas reacts with 2 moles of water vapor.
Therefore, the amount of moles of oxygen gas required is half of the moles of water vapor produced.
Moles of oxygen gas = (1/2) * (30.99 moles) = 15.495 moles

Step 4: Calculate the grams of oxygen gas using the molar mass of oxygen gas.
The molar mass of oxygen gas (O2) is about 32 g/mol.
Grams of oxygen gas = Moles of oxygen gas * Molar mass of oxygen gas
Grams of oxygen gas = 15.495 moles * 32 g/mol
Grams of oxygen gas = 495.84 g

Therefore, approximately 495.84 grams of oxygen gas is required to produce 7.46 kJ of heat.

To answer this question, we need to use the balanced chemical equation and the heat of reaction given.

Given:
ΔH = -484 kJ (heat of reaction for the combustion of hydrogen gas)
We need to find the number of grams of oxygen gas required to produce 7.46 kJ of heat.

The balanced chemical equation is:
2 H₂(g) + O₂(g) → 2 H₂O(g)

From the equation, we can see that the ratio of the coefficients is 2:1 for hydrogen gas to oxygen gas. This means that for every 2 moles of hydrogen gas that react, 1 mole of oxygen gas is required.

First, let's calculate the moles of heat released when 7.46 kJ of heat is released:
ΔH = -484 kJ
Number of moles of heat released = ΔH / ΔH of the reaction
Number of moles of heat released = 7.46 kJ / (-484 kJ)

Next, since we know the ratio of moles of hydrogen gas to oxygen gas is 2:1, we can use this ratio to calculate the moles of oxygen gas required.

Number of moles of oxygen gas required = (Number of moles of heat released) / 2

Now that we have the moles of oxygen gas required, we can calculate the grams of oxygen gas required using the molar mass of oxygen:

Grams of oxygen gas required = (Number of moles of oxygen gas required) × (Molar mass of oxygen)

The molar mass of oxygen (O₂) is 32.00 g/mol.

Now you can substitute the values and calculate the grams of oxygen gas required.

484 kJ heat released from 4 g hydrogen.

44.0 kJ heat used to vaporize the water/mole or 88 kJ heat used to vaporize 4 g hydrogen.
484-88 = 396 kJ net heat released.
So We need 484 x (x grams/4) = 7.46 g hydrogen burned to produce the 396 kJ.
Solve for x THEN CHECK IT TO MAKE SURE THE REACTION RELEASES ENOUGH TO USE UP THE AMOUNT NEEDED TO VAPORIZE AND LEAVE YOU WITH 7.45 kJ.
Check my thinking.