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Calculate the number of grams of magnesium nitride necessary to produce 2.02 L of ammonia gas at STP.

I got 4.55g for the upper question which is correct but idk how to do the bottom one:

How many moles of magnesium hydroxide can also be formed?

Mg3N2(s) + 6 H2O(l)--> 3 Mg(OH)2(aq) + 2 NH3(g)


  • chem -

    I find it amazing you can do the first part but not the second part. Both are the same kind of stoichiometry problem. Here is the first one, then to the second one. I want you to see the similarity.
    2.02L at STP = 2.02L x (1 mol/22.4L) = 0.09018 moles NH3.
    Using the coefficients in the balanced equation, convert moles NH3 to moles Mg3N2. 0.09018 moles NH3 x (1 mole Mg3N2/2 moles NH3) = 0.09018 x (1/2) = 0.04509 moles Mg3N2.
    Now convert moles Mg3N2 to grams. g = moles x molar mass = 0.04509 x 100.93 = 4.55 g Mg3N2.

    Since we had to determine moles NH3, we may as well start there. moles NH3 = 0.09018.
    Using the coefficients in the balanced equation convert moles NH3 to moles Mg(OH)2. moles NH3 x (3 moles Mg(OH)2/2 moles NH3) = 0.09018 x (3/2) = 0.1357 moles Mg(OH)2.
    Convert moles to grams by g = moles x molar mass. 0.1357 moles Mg(OH)2 x molar mass Mg(OH)2 = ??

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