3. Balance the chemical equations​

A) BaCl2 + 2NaOH → Ba(OH)2 + 2NaCl

B) H2SO4 + 2KOH → K2SO4 + H2O

C) K2O3 + 2HCl → 2KCl + H2O + CO2

D) 2FeSO4 → Fe2O3 + SO2 + SO3

E) 2NaNO3 → 2 NaNO2 + O2

Explanation:

A) BaCl² + 2NaOH → _Ba(OH)² + 2NaCl

B) H²SO⁴ + 2KOH → _K²SO⁴ + H²O

C) K²CO³ + 2HCl → _KCl + H²O + CO²

D) 2FeSO⁴ → _Fe²O³ + _SO² + _SO³

E) 2NaNO³ → 2NaNO² + _O²

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.