which laboratory process is used to determine the concentration of one solution by using a volume of another solution of known c
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45 moles of H₂ are theoretically needed to produce 30.0 moles of NH₃
<h3>Further explanation</h3>Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.
- Mole
The mole itself is the number of particles contained in a substance amounting to 6.02.10²³
Mole can also be sought if the amount of substance mass and its molar mass is known
Reaction that happens :
N₂ +3H₂ ⇒ 2NH₃
mole N₂ : H₂ : NH₃ = 1 : 3 : 2
To produce 30.0 moles of NH₃,
- H₂ needed :
mole H₂ = 45 mole
- N₂ needed :
mole N₂ = 15 mole
So the minimum N₂ needed is: 15 mole
14.5 moles of N₂ can only produce NH₃ :
mole NH₃ = 29 mole
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Keywords: mole, NH₃, N₂, H₂
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.