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8 March, 02:58

In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 > 4NO + 6H2O a. Which reactant is the limiting reagent? O2 b. How many grams of NO are formed

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  1. 8 March, 04:10
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    2.61 g of NO will be formed

    The limiting reagent is the O₂

    Explanation:

    The reaction is:

    4NH₃ + 5O₂ → 4NO + 6H₂O

    We convert the mass of the reactants to moles:

    3.25g / 17 g/mol = 0.191 moles of NH₃

    3.50g / 32 g/mol = 0.109 moles of O₂

    Let's determine the limiting reactant by stoichiometry:

    4 moles of ammonia react with 5 moles of oxygen

    Then, 0.191 moles of ammonia will react with (0.191. 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

    Ratio with NO is 5:4

    5 moles of oxygen produce 4 moles of NO

    0.109 moles will produce (0.109. 4) / 5 = 0.0872 moles of NO

    We convert the moles to mass, to get the answer

    0.0872 mol. 30g / 1 mol = 2.61 g
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