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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Caden on October 14, 2014, 11:07:09 AM

Title: Magnesium nitride stoichiometry question
Post by: Caden on October 14, 2014, 11:07:09 AM

I've been trying to solve this problem:  If you react Mg in a crucible filled with air to form MgO and Mg₃N₂ you end with 0.47g
You then add water which reacts with the Magnesium nitride to form MgO and ammonia gas.  The total amount of MgO is 0.486g.
The balanced equations came out as
2Mg + O₂ :rarrow: 2MgO
3Mg+ N₂  :rarrow: Mg₃N₂
Mg₃N₂ + 3H₂O  :rarrow: 3MgO + 2NH₃
I need to figure out how much Mg₃N₂ I started with.
Beyond that I am stumped.  I converted the .486g of MgO into moles, but couldn't see anywhere to go with that.  I know there is some algebraic principles involved here but I don't know where to start.

Title: Re: Magnesium nitride stoichiometry question
Post by: Corribus on October 14, 2014, 11:38:38 AM

You need to say what you tried... and also you need to say what the actual question is. What does it want you to find? (I'm guessing, the starting amount of Mg.)

Title: Re: Magnesium nitride stoichiometry question
Post by: johnnyjohn993 on October 20, 2014, 06:47:46 AM

You have : 0.486g of MgO  Using stoichiometric relationship


0.486g MgO x ___1mol MgO____ x____1mol Mg₃N₂___x___100.95 g/mol Mg3N2_____
                      40.31 g/molMgO                       3mol MgO                               1molMg3N2             

=  0.406g of Mg₃N₂

Title: Re: Magnesium nitride stoichiometry question
Post by: mjc123 on October 20, 2014, 08:35:15 AM

So 0.486g MgO is equivalent to 0.406g Mg₃N₂.
But you had a mixture of the two solids, weighing 0.47g. How much Mg₃N₂ was in that mixture?