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#

1 Answer
Aug 2, 2017

Answer:

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#.