Ammonia is a principal nitrogen fertilizer. It is prepared...

Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen.$$3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$In a particular reaction, 6.0 moles of $\mathrm{NH}_{3}$ were produced. How many moles of $\mathrm{H}_{2}$ and how many moles of $\mathrm{N}_{2}$ were reacted to produce this amount of $\mathrm{NH}_{3} ?$

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Key Concepts

Balanced Chemical Equations

Balanced chemical equations are representations of chemical reactions where the number of atoms for each element is conserved on both sides of the equation. This balance is crucial because it reflects the law of conservation of mass. By ensuring that the equation is balanced, one can accurately use it to determine the mole ratios between reactants and products, which is the backbone of stoichiometric calculations.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of its constituent particles, such as atoms, molecules, or ions, via Avogadro's number. It enables chemists to count entities by weighing them, which is essential for performing quantitative analyses and converting between mass and number of particles.

Stoichiometry

Stoichiometry is the quantitative study of the reactants and products in a chemical reaction. It involves using the mole ratios derived from a balanced chemical equation to determine how much of one substance is needed or produced, given the amount of another substance. This concept is pivotal in predicting yields, calculating reagent quantities, and scaling reactions in both laboratory and industrial processes.

Transcript

00:12 So for this problem, we're given the following reaction of hydrogen with nitrogen gas with nitrogen gas to produce ammonia.

00:23 And we're told that in a specific reaction, 6 .0 moles of ammonia are formed.

00:35 And we need to determine how many moles of hydrogen and how many moles of nitrogen and how many moles of nitrogen were.

00:45 Used in order to produce the six moles of ammonia.

00:54 So to do this, we'll use the mole and mass conversion method.

00:59 So to solve problems like this, i always start with what we're given in the problem.

01:04 So that's, you know, we have six moles of ammonia.

01:18 Now we're going to want to figure out what's the equivalent of the number of moles of, let's say nitrogen.

01:29 We'll solve.

01:29 For the moles of nitrogen first.

01:34 So what we do is we look at the mole ratio between nitrogen and the ammonia.

01:39 So we see that for every two moles of ammonia produced, we only used, or the reaction, only uses one mole of nitrogen.

01:53 So we put the number of moles of ammonia on the bottom, two moles, ammonia for every one mole of nitrogen.

02:04 And, we put the number of ammonia on the bottom.

02:05 And we're solving for the number of moles...

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