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aleksandrvk [35]
3 years ago
12

Calculate the amount of mole of iron produced from the reaction of 15.9 grams of iron oxide.

Chemistry
1 answer:
Usimov [2.4K]3 years ago
3 0

Answer:

0.19875

Explanation:

nFe2O3=0.099375

nFe=2nFe2O3=0.19875

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Identify the oxidizing and reducing agents in the following: H2S(aq) + Cl2(g) -> S(s) + 2HCI (aq)
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The oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.

<h3>What is an oxidizing and reducing agent?</h3>

An oxidizing agent is any substance that oxidizes, or receives electrons from another substance and as a result, becoming reduced.

On the other hand, a reducing agent is any substance that reduces or donates electrons to another and as a result becomes oxidized.

According to this reaction; H2S(aq) + Cl2(g) -> S(s) + 2HCI (aq)

  • H2S accepts electrons from Cl2 and becomes reduced to S
  • Cl2 donates electrons to H2S and becomes oxidized to HCl

Therefore, the oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.

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2 years ago
Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of Ag_2O = 25.0 g

Mass of C_{10}H_{10}N_4SO_2 = 50.0 g

Molar mass of Ag_2O = 231.7 g/mole

Molar mass of C_{10}H_{10}N_4SO_2 = 250.3 g/mole

Molar mass of AgC_{10}H_{9}N_4SO_2 = 357.1 g/mole

First we have to calculate the moles of Ag_2O and C_{10}H_{10}N_4SO_2.

\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles

\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)

From the balanced reaction we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 1 mole of Ag_2O

So, 0.1998 moles of C_{10}H_{10}N_4SO_2 react with \frac{0.1998}{2}=0.0999 moles of Ag_2O

From this we conclude that, Ag_2O is an excess reagent because the given moles are greater than the required moles and C_{10}H_{10}N_4SO_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgC_{10}H_9N_4SO_2

From the reaction, we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 2 mole of AgC_{10}H_9N_4SO_2

So, 0.1998 mole of C_{10}H_{10}N_4SO_2 react with 0.1998 mole of AgC_{10}H_9N_4SO_2

Now we have to calculate the mass of AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

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Answer:

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This radiation is the one used in microwaves.

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