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7 November, 02:58

How many grams of ammonia must you start with to make 900.00 l of a 0.140 m aqueous solution of nitric acid? assume all the reactions give 100% yield?

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  1. 7 November, 03:04
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    You need the set of reactions that goes from ammonia to nitric acid.

    1) 4NH3 (g) + 5O2 (g) - - >4NO (g) + 6H2O (g)

    2) 2NO (g) + O2 (g) - - >2NO2 (g)

    3) 3NO2 (g) + H2O (l) - - >2HNO3 (aq) + NO (g)

    State the ratio of moles of HNO3 to NH3:

    4 moles of NH3 produce 4 mole of NO,

    4 moles of NO produce 4 moles of NO2

    4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.

    => (8/3) moles HNO3 : 4 moles NH3

    Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution

    M = n / V = > n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3

    Use proportions:

    (8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x

    => x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3

    Convert moles to grams:

    molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol

    mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g

    Answer: 3213 g.
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