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snow_tiger [21]
2 years ago
15

How many molecules of Mg3N2 (magnesium nitride) are formed when excess Mg (magnesium)

Chemistry
2 answers:
dybincka [34]2 years ago
7 0

Explanation:

<em>3Mg(s) + N2(g) = Mg3N2(s)</em>

First check that the equation is balanced. In this case, it is.

Assuming that magnesium is the limiting reactant:

  1. First find the molecular weight using the Periodic Table.

       We find that the atomic mass of magnesium is approximately

       <em>24.3g</em>, so the molecular weight is just <em>24.3g\mol</em>

   

    2. Next we need the mole to mole ratio. As there are <em>3</em>

        magnesiums for <em>1</em> magnesium nitride (shown by the coefficients), the                    

        mole to mole ratio is<em> 1 mol Mg3N2\3 mol Mg.</em>

   

    3. We need the amount of the substance, in grams. Since you have not    

        stated it in the question, I'll just do <em>10g</em> AS AN EXAMPLE. Note that    

       depending on the amount, the LIMITING REAGENT MAY DIFFER.

   4.  Finally, we need the molecular weight of <em>Mg3N2</em>, which we can easily    

        calculate to be around <em>100.9\mol.</em>

<em />

   5.  Putting this all together, we have<em> 10gMg⋅ (mol Mg\24.3gMg) </em>

<em>         (1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)</em>

     

        the units will cancel to leave <em>gMg3N2</em> (grams of magnesium nitride):

       

<em>        10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)</em>

<em>        (100.9g Mg3N2\mol Mg3N2)</em>

<em />

Doing the calculation yields approximately 13.84g.

Assuming that nitrogen is the limiting reactant:

Similarly, following the above steps but with <em>10g</em> of nitrogen yields <em>36.04g</em>

In conclusion, as we produce less amount of <em>Mg3N2</em> when we assumed that <em>Mg</em> was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

Note: This is in THIS CASE, where we have <em>10g</em> of both. The answer may vary depending on the amount of each substance.

Mars2501 [29]2 years ago
6 0

Answer:

  1.46×10^25 molecules

Explanation:

Each mole of N₂ reacts with 3 moles of Mg to form 1 mole of Mg₃N₂. That is, the number of moles of product is equal to the number of moles of nitrogen gas when excess magnesium is present.

There are 6.022×10²³ molecules in each mole of a substance, so the number of molecules of Mg₃N₂ is

  (24.3 mol)(6.022×10²³ molecules/mol) = 1.46×10²⁵ molecules

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