When 100 g #Mg_3N_2# reacts with 75.0 g #H_2O#, what is the limiting reactant?

1 Answer
anor277
May 7, 2018

The salt REQUIRES SIX EQUIV of water....and the water is the limiting reactant....

Explanation:

As always, our priority is to write a stoichiometrically balanced equation to represent the reaction....

#underbrace(Mg_3N_2(s) + 6H_2O(l))_"209 g" rarr underbrace(3Mg(OH)_2(aq) + 2NH_3(aq))_"209 g"#

I subscripted the masses of reactants and product to establish that the equation was stoichiometrically kosher. Garbage out EQUALS garbage in, as is absolutely required.

And so ...........
#"moles of magnesium salt"=(100*g)/(100.95*g*mol^-1)=0.991*mol#

#"moles of water"=(75.0*g)/(18.01*g*mol^-1)=4.16*mol#

And so here UNUSUALLY the water is in stoichiometric deficiency. And it is unusual in that in these circumstances it is hard to exclude water from a reaction setup. The nitride may be incompletely reduced to give say hydrazine or some other product.

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