Animated Solution for Chemistry - Basic Concepts in Chemistry: For a reaction, $N_2(g)+3H_2(g)⟶2N{H}_3(g)$, identify dihydrogen ($H_2$) as a limiting reagent in the following reaction mixtures.

Imagine you are hosting a party and making sandwiches. You have $10$ slices of bread but only $2$ slices of cheese. Even though you have plenty of bread, the cheese limits you to making only $2$ sandwiches. In chemistry, the reactant that runs out first is called the limiting reagent. It dictates exactly how much product can be formed, regardless of how much of the other reactants you have lying around.

In this problem, we are tasked with finding out in which mixture hydrogen ($H_2$) acts as the limiting reagent for the synthesis of ammonia.

Before we test the mixtures, we must understand the perfect, ideal recipe for making ammonia. We start with the balanced chemical equation:

Let's translate this molar ratio into a mass ratio. The molar mass of nitrogen gas ($N_2$) is $28\text{ g/mol}$, and the molar mass of hydrogen gas ($H_2$) is $2\text{ g/mol}$. According to the balanced equation, $1$ mole of $N_2$ reacts with $3$ moles of $H_2$.

Therefore, the standard mass requirement is:

This means that for every $28\text{ g}$ of nitrogen, you need exactly $6\text{ g}$ of hydrogen for a complete reaction with nothing left over.

Now, let's evaluate the first mixture given in option (a): $56\text{ g of }{N}_2$ and $10\text{ g of }{H}_2$.

To find out who runs out first, we ask a simple question: If we want to consume all $56\text{ g}$ of nitrogen, how much hydrogen do we actually need? We can use the unitary method based on our standard stoichiometry.

Notice how $56$ is exactly double of $28$. This makes the math beautifully simple. We need exactly double the hydrogen:

Here is the catch! We mathematically require $12\text{ g}$ of hydrogen to completely react with the $56\text{ g}$ of nitrogen. However, if we look at the mixture provided in option (a), we only have $10\text{ g}$ of hydrogen available.

Since the available hydrogen ($10\text{ g}$) is strictly less than the required hydrogen ($12\text{ g}$), the hydrogen will be completely consumed before the nitrogen runs out.

If we quickly glance at the other options, we find that in option (b), $35\text{ g}$ of $N_2$ only needs $7.5\text{ g}$ of $H_2$ (we have $8\text{ g}$, so $H_2$ is in excess). In option (c), $14\text{ g}$ of $N_2$ needs $3\text{ g}$ of $H_2$ (we have $4\text{ g}$, excess again). And in option (d), $28\text{ g}$ of $N_2$ perfectly matches the $6\text{ g}$ of $H_2$, meaning neither is limiting. Thus, option (a) is the definitive correct answer.