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22 March, 23:05

If 22.5 liters of oxygen reacted with excess of hydrogen, how many liters of water vapor could be produced?

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Answers (2)
  1. 22 March, 23:14
    0
    2H2 + O2 / rightarrow 2H2O

    PV = nRT

    since assuming that the reactants and product are all at the same temperature and pressure, the molar ratio of reactants = volume ratio of reactants.

    The chemical reaction between Hydrogen and oxygen is as follows:

    2H2 + O2 - --> 2H2O

    2 ... 1 ... 2 (volume ratio = molar ratio at same condition of temperature and pressure)

    Given that

    oxygen volume = 22.5L

    Volume of H2O = 22.5 L volume oxygen * 2 volume H2O / 1.0 L volume oxygen

    = 45 L water
  2. 22 March, 23:25
    0
    45 L

    Explanation:

    Assuming normal temperature and pressure (NTP) and ideal gas behaviour, 22.5 L are equivalent to:

    PV = nRT

    n = PV/RT

    n = (1*22.5) / (0.082*273.15)

    n = 1 mol

    Oxygen and hydrogen react to form water as follows:

    2 H2 (g) + O2 (g) - > 2 H2O (g)

    From the balanced equation, we know that 1 mol of Oxygen produces 2 moles of water. The volume occupied by these 2 moles is:

    PV = nRT

    V = nRT/P

    V = 2*0.082*273.15/1

    V = 45 L

    Know we can see the assumption of NTP conditions was unnecessary, the ratio between moles of oxygen consumed and moles of water produced is 1:2, so the ratio between volumes.
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