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1 February, 19:42

Balance the chemical equation given below, and calculate the volume of nitrogen monoxide gas produced when 8.00 g of ammonia is reacted with 12.0 g of oxygen at 25°C? The density of nitrogen monoxide at 25°C is 1.23 g/L.___ NH3 (g) + ___ O2 (g) ? ___ NO (g) + ___ H2O (l) A) 4.88LB) 7.38LC) 11.5LD) 17.3LE) 19.2L

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  1. 1 February, 19:46
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    The volume of NO is 7.32 L (option B 7.28L is the closest to this)

    Explanation:

    Step 1: Data

    Mass of ammonia = 8.00 grams

    Mass of oxygen = 12.0 grams

    Temperature = 25.0 °C

    Density of NO = 1.23 g/L

    Step 2: The balanced equation

    4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)

    Step 3: Calculate moles ammonia

    Moles ammonia = 8.00 grams / 17.03 g/mol

    Moles ammonia = 0.470 moles

    Step 4: Calculate moles oxygen

    Moles O2 = 12.0 grams / 32.0 g/mol

    Moles O2 = 0.375 moles

    Step 5: Calculate limiting reactant

    For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

    O2 is the limiting reactant. It will completely be consumed (0.375 moles).

    NH3 is in excess. There react 4/5*0.375 moles = 0.300 moles

    There remain 0.470 - 0.300 = 0.170 moles NH3

    Step 6: Calculate moles NO

    For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

    For 0.375 moles O2 we'll have 0.300 moles NO produced

    Step 7: calculate mass NO

    Mass NO = moles * molar mass

    Mass NO = 0.300 moles * 30.01 g/mol

    Mass NO = 9.003 grams

    Step 8: Calculate volume NO

    Volume NO = mass NO / density

    Volume NO = 9.003 grams / 1.23 g/L

    Volume NO = 7.32 L

    The volume of NO is 7.32 L (option B 7.28L is the closest to this)
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