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mihalych1998 [28]
3 years ago
9

How many moles of lithium chloride will be formed by the reaction of chloride with 4.28 mol of lithium bromide in the following

reaction (below)? *

Chemistry
1 answer:
lys-0071 [83]3 years ago
7 0

Answer:

The answer is 4.28 moles

Explanation:

This is super easy okay, you won't forget this!

Basically mole ratios, we're just looking at the coefficients in front of the compounds, multiplying them, and dividing them as we see fit.

In this example, you can see how you need 2 moles of lithium bromide (LiBr) for the reaction, and 2 moles of lithium chloride (LiCl) will be produced.

Basically, the <u>molar ratio</u> is when you divide numbers and see how much of this do I have for that (if that makes sense).

So if you were to divide the 2 moles of LiBr / 2 moles of LiCl = 1. So we know that the mole ratio for LiBr to LiCl is 1:1 or 2:2, either or, it's the same thing.

SO THE BIG IDEA, if we have 4.28 moles of lithium bromide reacting, we should also have 4.28 moles of lithium chloride produced, BECAUSE the <u>mole ratio</u> is 1:1.

I hope this makes sense please tell me if it doesn't, I will try my best to explain a little more.

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"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

3 0
4 years ago
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