According to the balanced chemical equation, 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor.
Therefore, to produce 2 moles of water vapor, we need 2 moles of hydrogen gas.
From the ideal gas law, 1 mole of gas occupies 22.4 L at standard temperature and pressure.
So, 2 moles of hydrogen gas will occupy 2 x 22.4 = 44.8 L.
Therefore, to produce 8 L of water vapor, we will need 8/44.8 = 0.1786 moles of hydrogen gas.
However, as per the stoichiometry of the reaction, 2 moles of hydrogen gas are needed to produce 2 moles of water vapor.
Hence, to produce 0.1786 moles of water vapor, we would need 0.1786 moles of hydrogen gas.
Therefore, about 0.1786 moles of hydrogen gas will be needed to make 8 L of water vapor.
Determine the volume of hydrogen gas needed to make 8L of water vapor. 2H2(g)+O2(g) -> 2H2O(g)
1 answer