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Andrei [34K]
2 years ago
7

A student wants to prove the concept of limiting reactants to his lab group during class. If the student adds 56 g or Fe to 71 g

of Cl2, what will the likely result of the reaction be in regards to limiting/excess reactants?
Chemistry
1 answer:
Alexeev081 [22]2 years ago
7 0

The result of the reaction will be such that Cl2 is limiting in its availability

<h3>Limiting reactants</h3>

From the equation of the reaction:

2Fe + 3Cl2 ---> 2FeCl3

Mole ratio of Fe and Cl2 = 2:3

Mole of 56 g of Fe = 56/56 = 1

Mole of 71 g of Cl2 = 71/71 = 1

Whereas, 1 mole of Fe requires 1.5 moles of Cl2. Thus, Cl2 is limiting in availability.

More on limiting reactants can be found here: brainly.com/question/14225536

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Consider the reaction of A(g) + B(g) + C(g) =&gt; D(g) for which the following data were obtained:
Drupady [299]

Answer:

r = k [A]^{2}[B]^{2}

Explanation:

A + B + C ⟶ D

\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}

5 0
3 years ago
Please help me with this? :(
Harrizon [31]

Answer:

1. Diagram C.

2. Diagram A.

Explanation:

1. Calcium atom, Ca has 20 protons and 20 electrons. On the other hand, Calcium ion, Ca^2+ has 20 protons and 18 electrons. This is true because the +2 charge on the calcium ion, Ca^2+ indicates that the calcium atom, Ca has loss 2 electrons.

From the above illustration we can say that calcium ion, Ca^2+ has the following:

Proton = 20

Electron = 18

Therefore, diagram C indicates calcium ion, Ca^2+.

2. Fluorine atom, F has 9 protons and 9 electrons. Fluoride ion, F¯ has 9 protons and 10 electrons. This is so because the –1 charge on the fluoride ion, F¯ indicates that the fluorine atom, F has gained 1 electron.

Thus, we can say that the fluoride ion, F¯ has the following:

Proton = 9

Electron = 10

Therefore, diagram A represent fluoride ion, F¯.

8 0
3 years ago
Often in organic synthesis reactions, multiple products will be synthesized that need to be separated from each other. The separ
stiv31 [10]

Answer:

True

Explanation:

The chemical method of separating two compounds depends on their difference in solubility in water, ether, dilute acid or alkali. Separatory funnel thus, are used to distinguish immiscible fluids from their solutes. The funnel is usually made of glass, pear-shaped, and usually contains a stopper and a stopcock.

8 0
3 years ago
Why series fatty acid carbons seen in double from?
BARSIC [14]
Stearic acid is a fully saturated fatty acid with no carbon-carbon double bonds. (Bottom) Oleic acid is an unsaturated fatty acid.
8 0
2 years ago
Give five examples of structures with this formula (c6h12). at least one should contain a ring, and at least one should contain
hammer [34]

Answer: -

Following are five examples of structures with the chemical formula C₆H₁₂

Compound A is Hexene.

Compound B is 2-Hexene.

Compound C is 3-Hexene.

Compound D is Cyclohexane.

Compound E is Methylcyclopentane.

As we can see Hexene, 2- Hexene and 3-Hexene all have double bonds.

Cyclohexane and Methylcyclopentane contains a ring.

7 0
3 years ago
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