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shusha [124]
3 years ago
6

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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A rock with different rocks combined to make it

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7 0
3 years ago
Read 2 more answers
Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
Ghella [55]

Answer:

(a) ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = -2.881 J/K; total change in entropy = 0

(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

(c) ΔS_{sys}  = 0 ; ΔS_{sur}  = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

6 0
3 years ago
The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation
Reil [10]

Answer:

1027.62 g

Explanation:

For B_2H_6  :-

Mass of B_2H_6  = 296.1 g

Molar mass of B_2H_6  = 27.66 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{296.1\ g}{27.66\ g/mol}

Moles\ of\ B_2H_6= 10.705\ mol

From the balanced reaction:-

B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}

1 mole of B_2H_6 react with 3 moles of oxygen

Thus,

10.705 mole of B_2H_6 react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

8 0
3 years ago
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maks197457 [2]

Answer:

A) a plant

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