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.
Well, if we're talking about burning hydrogen gas to produce water, it sounds like we're about to have a hot mess! But let's calculate the grams of oxygen gas needed to produce that heat.
First, we need to convert the heat released during the reaction to kilojoules:
7.46 kJ = 7.46 * 10^3 J
Next, we can use the balanced equation to find the ratio of oxygen to heat released:
2 H2(g) + O2(g) → 2 H2O(g) ΔH = −484 kJ
From the equation, we see that the molar ratio of O2:H2 is 1:2, and the heat released is 484 kJ for 2 moles of H2 reacted. So, for 7.46 kJ of heat released, we need a proportionate amount of oxygen gas.
Let's set up a proportion:
(7.46 * 10^3 J / -484 kJ) = (x grams / 1 mole)
Cross-multiplying:
-484 kJ * x grams = 7.46 * 10^3 J * 1 mole
Simplifying:
x grams = (7.46 * 10^3 J * 1 mole) / -484 kJ
Now, we need to convert joules to kilojoules and moles to grams:
x grams = (7.46 * 10^3 * 10^-3 kg * 1 mole) / -484 * 10^3 J
Finally, we can do the math:
x grams = -15.41 kg
Wait a minute, a negative mass? That doesn't seem right! Maybe I made a calculation error somewhere...or perhaps this question is just trying to fuel my comedic side!
To determine the grams of oxygen gas required to produce 7.46 kJ of heat, we need to use the stoichiometry of the balanced chemical equation and the enthalpy change of the reaction.
The balanced equation is:
2 H2(g) + O2(g) → 2 H2O(g)
The enthalpy change of the reaction (ΔH) is -484 kJ.
First, let's calculate the moles of water produced:
Since the ratio between H2 and H2O is 1:1, the moles of water produced will be the same as the moles of hydrogen used in the reaction.
ΔH = -484 kJ
Moles of hydrogen gas = Moles of water = 7.46 kJ / -484 kJ/mol = -0.0154 mol
Next, since the ratio between O2 and water is 1:2, the moles of oxygen gas required will be twice the moles of hydrogen:
Moles of oxygen gas = 2 x Moles of hydrogen = 2 x (-0.0154 mol) = -0.0308 mol
Now, let's convert the moles of oxygen gas to grams. We'll need to use the molar mass of oxygen gas, which is 32 g/mol:
Grams of oxygen gas = Moles of oxygen gas x Molar mass of oxygen gas
Grams of oxygen gas = -0.0308 mol x 32 g/mol ≈ -0.986 g
Note: The negative sign indicates that the reaction consumes oxygen gas.
Therefore, approximately 0.986 grams of oxygen gas are required to produce 7.46 kJ of heat when hydrogen gas burns at constant pressure to produce gaseous water.
To determine the number of grams of oxygen gas required to produce 7.46 kJ of heat, we need to use the balanced chemical equation and the enthalpy change (ΔH) for the reaction.
The given balanced chemical equation is:
2 H2(g) + O2(g) → 2 H2O(g)
The enthalpy change (ΔH) for the reaction is -484 kJ.
From the balanced chemical equation, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. Therefore, the molar ratio between H2 and O2 is 2:1.
First, we need to calculate the number of moles of H2 that react using the heat of vaporization of liquid water. Since 44.0 kJ is required to vaporize 1 mole of liquid water, we divide 7.46 kJ by 44.0 kJ/mol to find the number of moles of water produced in the reaction.
7.46 kJ / 44.0 kJ/mol = 0.169 mol
This means that 0.169 moles of water are produced in the reaction.
Since the molar ratio between H2 and O2 is 2:1, we need twice the number of moles of oxygen gas to react. Therefore, we multiply the number of moles of water by 1/2 to get the number of moles of oxygen gas.
0.169 mol * 1/2 = 0.0845 mol
Now, we need to calculate the mass of oxygen gas in grams. To do this, we need to use the molar mass of oxygen, which is approximately 32 g/mol.
Mass = number of moles * molar mass
Mass = 0.0845 mol * 32 g/mol
Mass = 2.704 g
Therefore, approximately 2.704 grams of oxygen gas are required to produce 7.46 kJ of heat when hydrogen gas burns at constant pressure to produce gaseous water.