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21 December, 13:43

determine mass of water formed when 12.5 L NH3 (at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)

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  1. 21 December, 14:13
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    The mass of water formed is

    calculation

    Use the ideal gas equation to calculate the moles of NH3 and O2

    that is Pv = n RT

    where; P = pressure,

    V = volume,

    n = number of moles,

    R=gas constant = 0.0821 l. atm / mol. K

    make n the formula of the subject by diving both side by RT

    n = PV / RT

    The moles of NH3

    n = (1.50 atm x 12.5 L) / (0.0821 L. atm / mol. k x 298 K) = 0.766 moles

    The moles of O2

    = (1.1 atm x 18.9 L) / (0.0821 L. atm / mol. k x 323 K) = 0.784 moles

    write the reaction between NH3 and O2

    4 NH3 + 5 O2 →4 No + 6H2O

    from equation above 0.766 moles of NH3 reacted to produce

    0.766 x 6/4 = 1.149 moles of H2O

    0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20

    since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced = 0.9408 moles

    mass of H2O = moles x molar mass

    from periodic table the molar mass of H2O = (1 x2) + 16 = 18 g/mol

    mass = 18 g/mol x 0.9408 moles = 16.93 grams
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