Question

write the balanced chemical equation for the following reaction Potassium Chloride + barium iodide gives potassium iodide + Barium Chloride​

Answer 1

Answer:

Potassium Chloride react with Barium Iodide give Potassium Iodide and Barium Chloride .

Chemically :

KCl + BaI₂  ⇒ KI + BaCl₂

Balanced chemical equation as :

Just multiply by 2 in KCl and by 2 in KI

2 KCl + BaI₂  ⇒ 2 KI + BaCl₂

In reactant side                               In product side

K atom = > 2                                     2

Cl atom = > 2                                    2

Ba atom = > 1                                     1

I atom    = >  2                                     2

Now every atoms are equal in both side reactant as well as product side .

i.e. 2 KCl + BaI₂  ⇒ 2 KI + BaCl₂ .

Answer 2

$\bold{\underline{\underline{Answer:}}}$

$\bold{2KCl\:+\:BaI_2\rightarrow\:2KI\:+\:BaCl_2}$

$\bold{\underline{\underline{Step\:-\:by\:-\:step\:explanation:}}}$

Potassium Chloride reacts with Barium Iodide to form Potassium Iodide and Barium Chloride.

$\bold{\underline{\underline{Skeletoon\:reaction:}}}$

$\bold{KCl\:+\:BaI_2\rightarrow\:KI\:+\:BaCl_2}$

$\bold{\underline{\underline{Balanced\:reaction:}}}$

To balance just multiply 2 with KCl in the rectant side, so the reaction seems as :-

$\bold{2KCl\:+\:BaI_2\rightarrow\:KI\:+\:BaCl_2}$

Now multiply 2 with Kl in the product side, the new equation :-

$\bold{2KCl\:+\:BaI_2\rightarrow\:2KI\:+\:BaCl_2}$

Now let's be assured that our reaction is balanced.

So first we will the number of atoms of the compounds on the rectant side :-

$\bold{\underline{\underline{Reactant\:side\:of\:the\:reaction:}}}$

$\bold{2KCl\:+\:BaI_2}$

$\bold{K}$ = 2

$\bold{Cl}$ = 2

$\bold{Ba}$ = 1

$\bold{I}$ = 2

Now let's check on the product side :-

$\bold{2KI\:+\:BaCl_2}$

$\bold{K}$ = 2

$\bold{Cl}$ = 2

$\bold{Ba}$ = 1

$\bold{I}$ = 2

The number of atoms in LHS (rectant side) and RHS (product side) is equal.

•°• Our equation is balanced.