Respuesta :
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 [tex]S_8[/tex].
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 [tex]S_8[/tex] = 3.0 g
Molar mass of Ag = 107.8 g/mole
Molar mass of [tex]S_8[/tex] = 256 g/mole
Molar mass of [tex]Ag_2S[/tex] = 247.8 g/mole
First we have to calculate the moles of Ag and [tex]S_8[/tex].
[tex]\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles[/tex]
[tex]\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)[/tex]
From the balanced reaction we conclude that
As, 16 mole of [tex]Ag[/tex] react with 1 mole of [tex]S_8[/tex]
So, 0.0278 moles of [tex]Ag[/tex] react with [tex]\frac{0.0278}{16}=0.00174[/tex] moles of [tex]S_8[/tex]
From this we conclude that, [tex]S_8[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Ag[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Ag_2S[/tex]
From the reaction, we conclude that
As, 16 mole of [tex]Ag[/tex] react to give 8 mole of [tex]Ag_2S[/tex]
So, 0.0278 moles of [tex]Ag[/tex] react to give [tex]\frac{0.0278}{16}\times 8=0.0139[/tex] moles of [tex]Ag_2S[/tex]
Now we have to calculate the mass of [tex]Ag_2S[/tex]
[tex]\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S[/tex]
[tex]\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g[/tex]
Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.
