Tthe volume of water vapor produced from the given amounts of oxygen gas and hydrogen gas, we can use stoichiometry and the ideal gas law at STP. The volume of water vapor produced is approximately 44.9 liters.
To calculate the volume of water vapor produced, we need to use stoichiometry and the ideal gas law.
Given:
32.0 g of oxygen gas
Temperature: STP (273 K)
Pressure: 1 atm
1. Convert the mass of oxygen gas to moles using its molar mass. The molar mass of oxygen gas (O2) is 32.00 g/mol, so:
moles of O2 = 32.0 g / 32.00 g/mol = 1.00 mol
2. Use the stoichiometry of the balanced chemical equation to determine the moles of water vapor produced. From the equation, we see that 1 mole of oxygen gas produces 2 moles of water vapor. Therefore:
moles of H2O = 2.00 mol
3. Use the ideal gas law to find the volume of water vapor. The ideal gas law equation is: PV = nRT, where P is the pressure, V is the volume, 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.
At STP, the pressure is 1 atm and the temperature is 273 K. Substituting the values into the ideal gas law equation:
1 atm * V = 2.00 mol * 0.0821 L·atm/(mol·K) * 273 K
V = 2.00 mol * 0.0821 L·atm/(mol·K) * 273 K / 1 atm
V ≈ 44.9 L
The volume of water vapor produced is approximately 44.9 liters.