If 987g of oxygen gas reacts completely with hydrogen gas, what volume of water vapor will be produced (in L) at 0.954 atm and 465 degree=es celcius?

Do the steps outlined before; i.e., write and balance the equation, convert mols O2 to mols H2O, then use PV = nRT to solve for volume.

To find the volume of water vapor produced, we need to use the ideal gas law equation: PV = nRT, where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L.atm/mol.K), and T is the temperature in Kelvin.

First, we need to calculate the number of moles of water vapor produced by the reaction.

From the reaction equation:
2H2 + O2 → 2H2O

Given that the reaction is complete, we can assume that all the hydrogen gas will react with the oxygen gas to form water vapor in a 2:1 ratio. Therefore, the moles of H2O produced will be half the moles of oxygen gas consumed.

To calculate the moles of oxygen gas:
molar mass of O2 = 32 g/mol
moles of O2 = mass of O2 / molar mass of O2
= 987 g / 32 g/mol
= 30.84 mol

Moles of H2O produced = 30.84 mol / 2
= 15.42 mol

Now, we need to convert the temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
= 465 °C + 273.15
= 738.15 K

Finally, we can calculate the volume of water vapor using the ideal gas law equation. However, we need to approximate the volume of the container in which the reaction is taking place. Let's assume it's 1 liter.

PV = nRT
(0.954 atm) * V(L) = (15.42 mol) * (0.0821 L.atm/mol.K) * (738.15 K)
0.954V = 959.615
V ≈ 1005.92 L

Therefore, approximately 1005.92 liters of water vapor would be produced at 0.954 atm and 465 degrees Celsius.