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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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? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2

Novay_Z [31]

Data:
$M_{concentrated} = 2.5\:mol$
$V_{concentrated} = ?$
$M_{dilute} = 0.50\:mol$
$V_{dilute} = 100\:mL\to0.100\:L$
<span>
Formula: Dilution Calculations

</span> $M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
<span>
Solving:

</span>
$M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
$2.5 * V_{concentrated} = 0.50 * 0.100$
$2.5V_{concentrated} = 0.05$
$V_{concentrated} = \frac{0.05}{2.5}$
$\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark$ <span>

</span>

3 0

3 years ago

Calculate the mass of calcium chloride that contains 3.20 x 1024 atoms of chlorine.

frez [133]

Answer:

294.87 gm CaCl_2

Explanation:

The computation of the mass of calcium chloride is shown below:

But before that following calculations need to be done

Number of moles of chlorine atom is

= 3.20 × 10^24 ÷ 6.022 × 10^23

= 5.314 moles

As we know that

1 mole CaCl_2 have the 2 moles of chlorine atoms

Now 5.341 mole chloride atoms would be

= 1 ÷ 2 × 5.314

= 2.657 moles

Now

Mass of CaCl_2 = Number of moles × molar mass of CaCl_2

= 2.657 moles × 110.98 g/mol

= 294.87 gm CaCl_2

7 0

3 years ago

Solid cadmium sulfide reacts with an aqueous solution of sulfuric acid . Express your answer as a balanced chemical equation. Id

max2010maxim [7]

Explanation:

When solid cadmium sulfide reacts with an aqueous solution of sulfuric acid then the reaction will be as follows.

$CdS(s) + H_{2}SO_{4}(aq) \rightarrow CdSO_{4}(aq) + H_{2}S(g)$

Hence, ionic equation for this reaction is as follows.

$CdS(s) + 2H^{+}(aq) + SO^{2-}_{4}(aq) \rightarrow Cd^{2+}(aq) + SO^{2-}_{4}(aq) + H_{2}S(g)$

Therefore, net ionic equation for this reaction is as follows.

$CdS(s) + 2H^{+}(aq) \rightarrow Cd^{2+}(aq) + H_{2}S(g)$

8 0

3 years ago

If the half-life of a radioactive element is 4 days, how long will it take for three- fourths of a sample of the element to deca

Jlenok [28]

Answer:

$\boxed{\text{8 da}}$

Explanation:

The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"

The half-life of the element is the time it takes for half of it to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

$\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}$

$\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}$

3 0

3 years ago

A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach

Bogdan [553]

Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ + I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

6 0

3 years ago