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3 years ago

Which reactant is limiting?Show work

Chemistry

1 answer:

fenix001 [56]3 years ago

5 0

Answer:

1. Iodine is the limiting reactant

2. $6molAlI_3$

Explanation:

1. Balanced chemcial equation:

$2Al+3I_2\rightarrow 2AlI_3$

2. Theoretical mole ratio

It is the ratio of the coefficients of the reactants in the balanced chemical equation:

$\dfrac{2molAl}{3molI_2}$

3. Actual ratio

It is ratio of the moles available to reat:

$\dfrac{9molAl}{9molI_2}$

4. Comparison

$\dfrac{9molAl}{9molI_2}>\dfrac{2molAl}{3molI_2}$

Then, there are more aluminum available than what is needed to react with the 9 moles of iodine, meaning that the aluminum is in excess and the iodine will react completely, being the latter the limiting reactant.

Conclusion: iodine is the limiting reactant.

5. How much aluminum iodide will be produced?

Use the theoretical mole ratio of aluminum iodide to iodide:

$\dfrac{2molAlI_3}{3molI_2}\times 9molI_2=6molAlI_3\leftarrow answer$

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3 years ago

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2 years ago

Read 2 more answers

Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag. Of the two metals, copper

KatRina [158]

Answer:

Answer is given below.

Explanation:

Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.

In cell representation, half cell present left to salt-bridge notation $(\parallel )$ is anodic system and another half cell present right to salt-bridge notation $(\parallel )$ is cathodic system.

So anode is Cu and cathode is Ag.

oxidation: $Cu-2e^{-}\rightarrow Cu^{2+}(aq.)$

[reduction: $Ag^{+}+e^{-}\rightarrow Ag$ ] $\times 2$

-----------------------------------------------------------------------------------------------

chemical equation: $Cu+2Ag^{+}(aq.)\rightarrow Cu^{2+}(aq.)+2Ag$

Oxidizing agent is that species which takes electron from another species. Here $Ag^{+}(aq.)$ takes electron from Cu. Hence $Ag^{+}(aq.)$ is the oxidizing agent.

Reducing agent is that species which gives electron to another species. Here Cu gives electron to $Ag^{+}(aq.)$ . Hence Cu is the reducing agent.

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4 years ago

3. Once magnesium (Mg) gives up electron(s) to become an ion, it will become

xeze [42]

Answer:

Option B. Cation that is smaller than the original atom.

Explanation:

Magnesium is a divalent element. This implies that magnesium can give up 2 electrons to become an ion (cation) as shown below:

Mg —> Mg²⁺ + 2e¯

Next, we shall write the electronic configuration of magnesium atom (Mg) and magnesium ion (Mg²⁺). This can be written as follow:

Mg (12) = 2, 8, 2

Mg²⁺ (10) = 2, 8

From the above illustration, we can see that the magnesium atom (Mg) has 3 shells while the magnesium ion (Mg²⁺) has 2 shells.

This simply means that the magnesium ion (Mg²⁺) i.e cation is smaller that the original magnesium atom (Mg).

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3 years ago