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3 years ago

15

Given the reaction: Ca + 2H2O → Ca(OH)2 + H2 What is the total number of moles of Ca needed to react completely with 4.0 moles o

f H2O?

2 answers:

DiKsa [7]3 years ago

6 0

<u>Answer:</u> The number of moles of calcium metal required is 2.0 moles.

<u>Explanation:</u>

We are given:

Moles of water reacted = 4.0 moles

The given chemical reaction follows:

$Ca+2H_2O\rightarrow Ca(OH)_2+H_2$

By Stoichiometry of the reaction:

2 moles of water is reacted with 1 mole of calcium metal

So, 4.0 moles of water will react with = $\frac{1}{2}\times 4.0=2.0mol$ of calcium metal

Hence, the number of moles of calcium metal required is 2.0 moles.

GREYUIT [131]3 years ago

4 0

Since the given equation is balanced, we know the Ca and the H2O are in a 1:2 ratio (to see the ratio, look at the coefficients in front of the molecules). Then, there are a few different ways to solve the problem. I am going to show you how to solve it using a ratio of moles of Ca/moles of H2O: x/1=4/2. First, you cross multiply to get 2x=4. Then, you divide both sides by 2 to get x=2 moles of Ca.

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Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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7nadin3 [17]

Answer:

<h2>The answer is 9 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

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mass = 216 g

volume = 24 cm³

The density is

$density = \frac{216}{24} \\$

We have the final answer as

<h3>9 g/cm³</h3>

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