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30 September, 03:45

2CH3CH2OH (l) → CH3CH2OCH2CH3 (l) + H2O (l)

If 28.0 g of NH3 and 168.8 g of ClF3 are allowed to react, what masses of each product would be recovered assuming complete reaction?

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  1. 30 September, 03:53
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    There will be produced 58.14 grams of Cl2; 22.97 grams of N2 and 98.45 grams of HF

    Explanation:

    Step 1: The balanced equation

    2ClF3 (g) + 2NH3 (g) → N2 (g) + Cl2 (g) + 6HF (g)

    Step 2: Data given

    Mass of NH3 = 28.0 grams

    Molar mass of NH3 = 17.03 grams

    Mass of ClF3 = 168.8 grams

    Molar mass ClF3 = 92.45 g/mol

    Molar mass of N2 = 28.01 g/mol

    Molar mass of Cl2 = 70.9 g/mol

    Molar mass of HF = 20.01 g/mol

    Step 3: Calculate moles

    Moles of NH3 = mass NH3 / Molar mass NH3

    Moles NH3 = 28.0 / 17.03 = 1.64 moles

    Moles ClF3 = mass ClF3 / Molar mass ClF3

    Moles ClF3 = 168.8 grams / 92.45 g/mol

    Moles ClF3 = 1.83 moles

    Step 4: Calculate limiting reactant

    For 2 moles of NH3 we need 2 moles of ClF3 to produce 1 mole of Cl2, 1 mole of N2 and 6 moles of HF

    NH3 is the limiting reactant. It will completely be consumed (1.64 moles).

    ClF3 is in excess. There will be consumed 1.64 moles. There will remain 1.83 - 1.64 = 0.19 moles of ClF3

    0.19 moles of ClF3 = 0.19 * 92.45 = 17.57 grams

    Step 5: Calculate moles of Cl2, N2 and HF

    For 2 moles of NH3 we need 2 moles of ClF3 to produce 1 mole of Cl2, 1 mole of N2 and 6 moles of HF

    For 1.64 moles of NH3 consumed, we will produce:

    1.64 / 2 = 0.82 moles of Cl2

    1.64/2 = 0.82 moles of N2

    1.64 * 3 = 4.92 moles of HF

    Step 6: Calculate mass of products

    Mass = moles * Molar mass

    Mass of Cl2 = 0.82 moles * 70.9 g/mol = 58.138 grams

    Mass of N2 = 0.82 moles * 28.01 g/mol = 22.97 grams

    Mass of HF = 4.92 moles * 20.01 g/mol = 98.45 grams
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