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.

Answer 1

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}=(3.0)/(107.8g/mole)=0.0278moles`

`\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=(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
`(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
`(0.0278)/(16)* 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* \text{ Molar mass of }Ag_2S`

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

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