Formula of precipitate HNO3(aq) + Ba(OH)2(aq) --
(s)
ZnCl2(aq) + KOH(aq) -
(s)
CuSO4(aq) + H2S(ag) -
(s)
H3PO4(aq) + Ca(OH)2(aq) +
(s)
Na2CO3(aq) + NH4Cl(aq) --
s)
Better look in picture

a non redox reaction occurs when there is not a transfer of electrons. for redox reactions, which are reduction-oxidation reactions, a change in the oxidation number occurs. elements that are reduced gain electrons, and elements that are oxidized lose electrons.

explanation:

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

the answer is c i took the test