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30 March, 12:12

The combustion of ammonia in the presence of excess oxygenyields NO2 (g) and H2O (g) : 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) The combustion of 43.9 mg of ammonia produces how many grams ofNO2 (g) ? A. 178B. 0.119C. 43.9D. 0.954E. 2.58

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  1. 30 March, 12:35
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    There will be produced 0.119 grams of NO2 (option B)

    Explanation:

    Step 1: Data given

    Mass of ammonia = 43.9 mg

    Oxygen is in excess

    Molar mass of NH3 = 17.03 g/mol

    Step 2: The balanced equation

    4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g)

    Step 3: Calculate moles NH3

    Moles NH3 = Mass NH3 / molar mass NH3

    Moles NH3 = 0.0439 g / 17.03 g/mol

    Moles NH3 = 0.00258 moles

    Step 4: Calculate moles of NO2

    For 4 moles NH3 we need 7 moles of O2 to produce 4 moles NO2 and 6 moles of H2O

    For 0.00258 moles of NH3 we'll have 0.00258 moles of NO2

    Step 5: Calculate mass of NO2

    Mass of NO2 = moles NO2 * molar mas NO2

    Mass NO2 = 0.00258 moles * 46 g/mol

    Mass NO2 = 0.119 grams

    There will be produced 0.119 grams of NO2 (option B)
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