Animated Solution for Chemistry - Basic Concepts in Chemistry: For the following reaction, the mass of water produced from $445\text{ g}$ of ${\text{C}}_{57}{\text{H}}_{110}{\text{O}}_6$ is : $2{\text{C}}_{57}{\text{H}}_{110}{\text{O}}_6(s)+163{\text{O}}_2(g)⟶114{\text{CO}}_2(g)+110{\text{H}}_2\text{O}(l)$

Imagine you are planning a massive road trip. You wouldn't just start driving without a map, right? In chemistry, our map is the balanced chemical equation. It tells us exactly how much of each ingredient we need and how much product we will get.

For our problem, we are given the combustion reaction of a giant organic molecule:

Our goal is to find out how much water is produced from a specific amount of the reactant. To do this, we first need to extract the core relationship from our "map". By looking at the stoichiometric coefficients, we can see that $2$ moles of our complex organic reactant will produce exactly $110$ moles of water. This molar ratio is the fundamental bridge connecting the two substances.

Now, the problem gives us the amount of reactant in grams, not moles. So, we need to translate our molar map into a mass map. This means calculating the molar masses.

Let's start with our giant molecule, ${\text{C}}_{57}{\text{H}}_{110}{\text{O}}_6$. I know this looks terrifying, but let's take a breath and break it down. Carbon has an atomic mass of $12$, hydrogen is $1$, and oxygen is $16$.

Since our balanced equation uses $2$ moles of this reactant, the total mass involved is:

Next, we do the same for water (${\text{H}}_2\text{O}$). The molar mass of water is a familiar $18\text{ g/mol}$. For $110$ moles, the total mass is:

We have successfully translated our map! We now know that $1780\text{ g}$ of the reactant produces $1980\text{ g}$ of water.

But the question asks about a specific scenario: what happens if we only have $445\text{ g}$ of the reactant? This is where the simple unitary method comes to our rescue.

If $1780\text{ g}$ yields $1980\text{ g}$, then $1\text{ g}$ yields $\frac{1980}{1780}\text{ g}$. Therefore, $445\text{ g}$ will yield:

Before you start doing long division, pause and look at the numbers. JEE problems often have hidden mathematical elegance. Notice that $1780$ is exactly four times $445$!

This beautiful cancellation makes our calculation incredibly simple:

And there we have it! By carefully following the stoichiometric map and keeping an eye out for mathematical shortcuts, we arrived at the final answer of $495\text{ g}$.