What is the maximum mass, in grams, of NH3 that can be produced by the reaction of 1.0 g of N2 and 1.0 g of H2 using the reaction below?
N2+H2→NH3 (not balanced)

A) 0.60 g
B) 5.7 g
C) 1.2 g
D) 2.9 g

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  1. I worked this for someone just a day or so ago but I can't find it. Here is a summary.
    This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
    N2 + 3H2 ==> 2NH3

    mols N2 = grams/molar mass = ?
    mols H2 = grams/molar mass = ?

    Using the coefficients in the balanced equation, convert mols N2 to mols NH3.

    Do the same and convert mols H2 to mols NH3.

    You can see that mols NH3 produced is not the same; one of the answers must be wrong. The correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR.
    Now use the LR value and convert that mols to grams.
    g NH3 = mols NH3 x molar mass NH3

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  2. DrBob222 Please let me know if this is correct

    N2 + 3 H2 → 2 NH3

    (1.0 g N2) / (28.01344 g N2/mol) = 0.03570 mol N2
    (1.0 g H2) / (2.01588 g H2/mol) = 0.4961 mol H2

    0.03570 mole of N2 would react completely with 0.03570 x (3/1) = 0.1071 mole of H2, but there is more H2 present than that, so H2 is in excess and N2 is the limiting reactant.

    (0.03570 mol N2) x (2 mol NH3 / 1 mol N2) x (17.03056 g NH3/mol) = 1.2 g NH3

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  3. That's my answer also.

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  4. the balanced equation is a 2 in front of the N2

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