Question

Write the balanced chemical equation for the following
Sodium $+$ Water $\to$Sodium Hydroxide $+$ Hydrogen gas
A.${H_2}O + 2N{a_2} \to 2NaOH + {H_2}$
B.$Na + {H_2}O \to NaOH + {H_2}$
C.$Na + {H_2}O \to NaOH + {H_2}$
D.$N{a_2}O + {H_2}O \to NaOH + H$
E.$2Na + 2{H_2}O \to 2NaOH + {H_2}$

Answer 1

We are given a reaction of sodium with water. This reacting forms sodium hydroxide, and Hydrogen as products. In a balanced chemical equation, the total mass on both reactant and product sides is equal, thus following the law of conservation of mass. In a balanced chemical equation, the atoms of all elements are equal on the reactant and product sides.

Complete step-by-step answer: Balancing an equation is an important step in chemistry. A chemical equation must be balanced because according to the law of conservation of mass, mass can neither be created nor destroyed. There should be no change in mass when reactants change into products. Also, a balanced equation tells us how much product we can expect from a given amount of reactant. There are certain steps we should follow while balancing a chemical equation
$\left( 1 \right)$ The unbalanced chemical equation must be written from the chemical formulae of reactants and products. Here sodium reacts water to form sodium hydroxide and hydrogen gas
$Na + {H_2}O \to NaOH + {H_2}$.
$\left( 2 \right)$ Now compare the atoms of each element on the reactant and product side. For example on the reactant side, there is two atoms of $H$ and on the product side there are three atoms of $H$
$\left( 3 \right)$ A stoichiometric coefficient describes the total number of molecules of a chemical species that participate in a chemical reaction. We get the total number of atoms participating in the reaction by multiplying the number of atoms of that particular element in the molecule with stoichiometric coefficient.The coefficient must balance the number of atoms on each side. For example, if we assign a stoichiometric coefficient $2$ to ${H_2}O$ and a stoichiometric coefficient $2$ to $NaOH$ , the total number of atoms of $H$ thus becomes balanced. We will notice that $4$ atoms of $H$ are there on the reactant side and product side. The equation becomes
$Na + 2{H_2}O \to 2NaOH + {H_2}$
$\left( 4 \right)$ Now, notice that there is one atom of $Na$ on the reactant side and two atoms of $Na$ on the product side. To make total number of atoms equal, we will assign stoichiometric coefficient $2$ to $Na$ . thus the balanced equation is
$2Na + 2{H_2}O \to 2NaOH + {H_2}$

Thus the correct option is $E$.

Note: It is important to note that they never try balancing a chemical equation by changing the subscript of the molecule. For example the subscript in ${H_2}O$ is $2$ . If we will change the subscript in a molecule, it will result in a new molecule. Thus a chemical equation is balanced by considering the stoichiometric coefficient in the equation.