Question

How much water is formed when $1.0$ mol of $HCl$ reacts completely with $1.0$ mol of $NaOH$?

Answer 1

In the field of science, a mole is characterized as the measure of a substance that contains precisely $6.02214076 \times {10^{23}}$'rudimentary elements' of the given substance. The number $6.02214076 \times {10^{23}}$ is prominently known as the Avogadro steady and is frequently signified by the image$'{N_A}'$.Originally it was defined as, the number of atoms in $12g$ of $carbon - 12$.

Complete step by step answer: Now, the definition is just one mole is $6.023 \times {10^{23}}$ of a chemical unit.
In a chemical calculation, the reactant is on the left side and the product is on the right side. But when it comes to the stable chemical equation, it is made known in the terms of moles.
Such that, how many moles are required in this reaction. Stoichiometry is the sturdy relationship of quantities of substances in a chemical reaction.
To estimate the number of moles, we need to use this equation;
Number of moles, $n = \dfrac{{mass}}{\text{molecular mass}}$
Assuming that $n$ is the number of moles, $m$ is the mass of substance, $M$ is the molar mass.
Looking at the equation:
$HCl + NaOH \Rightarrow {H_2}O + NaCl$.
Since all compounds in this reaction have $1$ mole, it will not affect the number of moles$\left( n \right)$.
If $1$ mole of $HCl$ or $NaOH$ gives you $1$ mole of ${H_2}O$, then the number of moles in ${H_2}O$ is:
$\left[ {\dfrac{1}{1} \times 1} \right]{\text{ }} = {\text{ }}1{\text{ }}mole$.
Now you have to find out the molar mass $\left( M \right)$ of water$\left( {{H_2}O} \right)$.
Before that, take note that you know what atoms are present in $\left( {{H_2}O} \right)$ - in that case, hydrogen and oxygens are present.
Refer to your periodic table and you can see molar mass of hydrogen and oxygen is:
$Hydrogen{\text{ }} = {\text{ }}1.0{\text{ }}g/mol$
$Oxygen{\text{ }} = {\text{ }}16.0{\text{ }}g/mol$
Therefore, the molar mass $\left( M \right)$ of $\left( {{H_2}O} \right)$ is:
$\left[ {2 \times 1.0 + 1 \times 16.0} \right]{\text{ }} = {\text{ }}18.0{\text{ }}g/mol$.
Last step, find the mass $\left( m \right)$ of water.
The formula of finding mass of substance $\left( m \right)$ is:
$m = n \times M$.
The mass of $\left( {{H_2}O} \right)$ is: $\left[ {m{\text{ }} = {\text{ }}1.0{\text{ }}moles{\text{ }} \times {\text{ }}18.0{\text{ }}\dfrac{{grams}}{{moles}}{\text{ }} = {\text{ }}18.0{\text{ }}grams} \right]$
Therefore $18.0{\text{ }}grams$ of water is molded when $1$ mole of $HCl$ reacts finally with $1$ mole of $NaOH$.

Note:
If a compound is having the mass, we can analyze the moles by using the formula of moles, either it is for one molecule or the compound. If a compound is having $A$ and $B$ molecules then the sum of the moles will be $1$.
${n_A} = \dfrac{\text{mass of A}}{\text{molecular mass of A}}$
${n_B} = \dfrac{\text{mass of B}}{{molecularmassofB}}$
${n_A} + {n_B} = 1$