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Identify the balanced chemical reaction between ZnCl${}_{2}$${}_{\ }$and NH${}_{4}$OH.
(A) $ZnCl_{2} +NH_{4} OH\to Zn(OH)_{2} \downarrow +NH_{4} Cl$
(B) $ZnCl_{2} +2NH_{4} OH\to Zn(OH)_{2} \downarrow +2NH_{4} Cl$
(C) $2ZnCl_{2} +2NH_{4} OH\to 2Zn(OH)_{2} \downarrow +NH_{4} Cl$
(D) $ZnCl_{2} +4NH_{4} OH\to Zn(OH)_{2} \downarrow +4NH_{4} Cl$


Answer
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Hint: A balanced equation in the chemical reaction is the one where the number of each element of the reactant is equal to the number of elements of the products. Also, the charge on each element of reactants is equal to charge on each element of the product.

Complete step by step answer:
To check whether the chemical equation is balanced or not, we first list the number of atoms of reactants and products
In reaction - $ZnCl_{2} +NH_{4} OH\to Zn(OH)_{2} \downarrow +NH_{4} Cl$


ReactantsProduct
$Zn=1$$Zn=1$
$Cl=2$$Cl=1$
$NH=4$$NH=4$
$OH=1$$OH=2$


As we see from the above table, the number of atoms of reactants is not equal to number of atoms of products
First, we see that the number of atoms of Cl in reactants is 2, whereas in products is 1. So let's multiply Cl of reactants with 2


ReactantsProduct
$Zn=1$$Zn=1$
$Cl=2$$Cl=1\times 2$
$NH=4$$NH=4$
$OH=1$$OH=2$


Therefore, we’ll multiply NH${}_{4}$Cl with 2 in order to balance chlorine (Cl) on both sides.
Now, we get equation :- $ZnCl_{2} +NH_{4} OH\to Zn(OH)_{2} \downarrow +2NH_{4} Cl$
The new number of reactants and products is as follows :


ReactantsProduct
$Zn=1$$Zn=1$
$Cl=2$$Cl=2$
$NH=4$$NH=8$
$OH=1$$OH=2$


Now we have an unbalanced number of NH on the reactants side, therefore we multiply NH${}_{4}$OH by 2 in order to balance NH on both sides.

We obtain equation :- $ZnCl_{2} +2NH_{4} OH\to Zn(OH)_{2} \downarrow +2NH_{4} Cl$

Now, the number of elements on both sides are as follows


ReactantsProduct
$Zn=1$$Zn=1$
$Cl=2$$Cl=2$
$NH=8$$NH=8$
$OH=2$$OH=2$


Thus, our equation is now balanced.

Therefore, our correct answer is B

Note: Make sure you only adjust the coefficients and not subscripts. For example in $Al+O_{2} \to Al_{2} O_{3} $${}_{\ }$ you cannot change the numbers 2 and 3 of oxygen, you can only multiply the elements in order to balance them. Like here we multiply Al with 4, oxygen with 3 and Aluminium Oxide with 2. We obtain equation :- $4Al+3O_{2} \to Al_{2} O_{3} $${}_{.}$