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sashaice [31]
3 years ago
5

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from

a mixture of 3.0g Ag and 3.0g.
Chemistry
1 answer:
IgorLugansk [536]3 years ago
4 0

The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g S_8.

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of S_8 = 3.0 g

Molar mass of Ag = 107.8 g/mole

Molar mass of S_8 = 256 g/mole

Molar mass of Ag_2S = 247.8 g/mole

First we have to calculate the moles of Ag and S_8.

\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles

\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)

From the balanced reaction we conclude that

As, 16 mole of Ag react with 1 mole of S_8

So, 0.0278 moles of Ag react with \frac{0.0278}{16}=0.00174 moles of S_8

From this we conclude that, S_8 is an excess reagent because the given moles are greater than the required moles and Ag is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag_2S

From the reaction, we conclude that

As, 16 mole of Ag react to give 8 mole of Ag_2S

So, 0.0278 moles of Ag react to give \frac{0.0278}{16}\times 8=0.0139 moles of Ag_2S

Now we have to calculate the mass of Ag_2S

\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S

\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

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