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3 years ago

What is meant by the limiting reactant why is it necessary to identify the limiting reactant when you want to know what?

Chemistry

1 answer:

WARRIOR [948]3 years ago

4 0

You have the reactants who react to make your products.

Reactants --------> Products

The limiting reactant is the reactant that will RUN OUT FIRST and that will establish the maximum amount of product that will be produced.

To make an example, let's look at this equation and at the following question:

2Al + 6HCl → 2AlCl3 + 3H2

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

We know that 2Al + 6HCL are the reactants, right?

These 2 reactants will react to create an amount of a certain product in this case they're asking us for the amount of H2 that will be produced. Therefore, we need to find the REACTANT it can be either AL OR HCL.

But, how do we know which one is the limiting reactant?

To know which one is the limiting reactant we need to know HOW MUCH H2 can Al produce, and also how much H2 can HCL produce. Why?

It's simple, because if you find the one that produces the less amount of H2, you know immediately that's the maximum amount of that product that will be produced.

Let's say you have a sandwich that needs to be made with 2 slices of bread, 3 meats and 1 cheese.

But you have got 4 slices of bread, 9 slices of meat, and 5 slices of cheese.

Well, you could make 2 sandwiches with 4 slices of bread, 3 sandwiches with those 9 slices of meat and 5 sandwiches with the 5 slices of cheese.

But, in reality, you can only make 2 sandwiches because you don't have any more bread to produce more sandwiches. You get it? That is the point of limiting reactants, to find what is the actual amount that can be produced.

Coming back to our equation, we can find the number of moles of H2 produced by each one of the reactants, Al and HCl. I'll find the number of moles quickly to show you what the concept of limiting reactant is.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As you can see, 6.5 mol of H2 is the maximum that can be produced by the HCL and is less than the Al, so that maximum amount that you will get in the product H2 is no more than 6.5.

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

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3 years ago

What is the oxidation number for iodine in Mg(IO3)2 ?

mestny [16]

The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5

6 0

3 years ago

Read 2 more answers

Calculate the molar mass of Al(CH3O2)3

lina2011 [118]

Answer:

The molar mass and molecular weight of Al(CH3CO2)3 is 204.1136.

Explanation:

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3 years ago

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Which of the following terms describes an accumulation of rocky, sandy, or clayey material deposited at the end of a glacier?

OLEGan [10]

Answer:

Option (D)

Explanation:

<u>Eskers are the long ridges that are comprised of rocks, sands and clay particles and are deposited towards the end of the glaciers</u>. These are fluvioglacial depositional features. These particles are exposed after the glaciers recede. These ridges are formed parallel to the earlier flow direction of ice. The size of eskers is generally smaller as it carries smaller particles such as rocks, sands, and gravels, in comparison to the different type of moraines. It is because the flow velocity decreases as the glaciers melt. So, these eskers are formed at the end of the glaciers.

Thus, the correct answer is option (D).

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Which of the following is not true of "concentration"?

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3 years ago