List the number of each type of atom on the right side of the equation hbr(aq)+2naoh(aq)→2nabr(s)+h2o(l)

On the basis of the given unbalanced equation, that is:

HBr (aq) + 2NaOH (aq) → 2NaBr (s) + H2O (l)

On the right side of the equation, there are 2 atoms of sodium (Na), 2 atoms of bromine (Br), 2 atoms of hydrogen (H), and 1 atom of oxygen (O₂).

After balancing the equation correctly we get:

HBr (aq) + NaOH (aq) → NaBr (s) + H2O (l)

On the right side, one atom of Na, 1 atom of Br, 1 atom of H and one atom of O₂.

As per the given unbalanced equation:-

HBr(aq) + 2NaOH(aq) --> 2NaBr(s) + H2O(l)

Na:-2

Br:-2

H:- 2

O:- 1

If correctly balanced then

HBr(aq) + NaOH(aq) --> NaBr(s) + H2O(l)

Na:-1

Br:-1

H:- 2

O:- 1

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.