Answer
2 N$H_{4}$Cl + Ba$(OH)_4$ $\longrightarrow$ 2N$H_{3}$ + Ba$Cl_{2}$ + 2 $H_{2}$O
Work Step by Step
N$H_{4}$Cl + Ba$(OH)_4$ $\longrightarrow$ N$H_{3}$ + Ba$Cl_{2}$ + $H_{2}$O
Writing the above equation for this chemical reaction is the first step. Using our knowledge on the nomenclature of names of ions and substances we obtain the above equation.
1- We start by balancing Cl atoms in both sides. Add a 2 before N$H_{4}$Cl as there is only one CL on the left and there are two on the right.
2 N$H_{4}$Cl + Ba$(OH)_4$ $\longrightarrow$ N$H_{3}$ + Ba$Cl_{2}$ + $H_{2}$O
2. Now we have 2 N on the left and only 1 on the right. So we add 2 before N$H_{3}$
2 N$H_{4}$Cl + Ba$(OH)_4$ $\longrightarrow$ 2 N$H_{3}$ + Ba$Cl_{2}$ + $H_{2}$O
3. Balance O atoms. there are 2 oxygen on the left, but only 1 on the right. So we add 2 before $H_{2}$O
2 N$H_{4}$Cl + Ba$(OH)_4$ $\longrightarrow$ 2 N$H_{3}$ + Ba$Cl_{2}$ + 2 $H_{2}$O.
Count the atoms of each elements and find our that we have balanced the equation.
