Question

How many moles of magnesium oxide are produced by the reaction of 3.82 g of magnesium nitride with 7.73 g of water? `Mg_3N_2 + 3H_2O -> 2NH_3 3MgO`

Answer 1

We assess the rxn....

`Mg_3N_2(s)+6H_2O(l) rarr 2NH_3(aq) + 3Mg(OH)_2(aq)`

Explanation:

Is it balanced? Don't trust my arithmetic.

This tells us UNEQUIVOCALLY that `100.95*g` `"magnesium nitride"` reacts with `108.06*g` water to give stoichiometric `"ammonia"`, and `"magnesium hydroxide"`.

We have molar quantities of........

`"moles of"` `Mg_3N_2=(3.82*g)/(100.95*g*mol^-1)=0.0378*mol`

`"moles of"` `OH_2=(7.73*g)/(18.01*g*mol^-1)=0.429*mol`

Water is thus present in stoichiometric excess. Do you agree?

And thus we get stoichiometric `"magnesium hydroxide"`, i.e. `0.0378*molxx3*molxx58.32*g*mol^-1=6.61*g`.