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)#

Is it balanced? My math isn't very good.

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 amounts of the following.

#"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#

Therefore, there is a stoichiometric excess of water present. Do you agree?

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

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Answer 2

Wyatt Adkins

0.233 moles of magnesium oxide are produced.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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 -&gt; 2NH_3 3MgO#

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#