Question

When a mixture of silver metal and sulfur is heated, silversulfide is formed: 16Ag(s) + S8(s)--> 8Ag2S(s)a.What mass of Ag2S is produced from a mixture of2.0 g of Ag and 2.0 g of S8?b.What mass of which reactant is left unreacted?

Answer 1

Mass of Ag₂S which is produced, 2.29 g

Mass of reactant in excess (S₈) which is left unreacted, 1.70g

Step-by-step explanation:

The balanced reaction is this:

16 Ag (s) + S₈ (s) → 8Ag₂S (s)

Molar mass of sulfur: 2g / 256.48 g/m = 0.00779 moles

Molar mass of silver: 2 g / 107.87 g/m = 0.0185 moles

For 16 moles of silver, I need 1 mol of S

For 0.0185 moles of Ag, I will need ( 0.0185 / 16) = 0.00116 moles

If I need 0.00116 moles of S, and I have 0.00779 moles it means, that S is my reactant in excess so the limiting reagent is the Ag.

Let's verify:

1 mol of S are needed to make react 16 moles of Ag

0.00779 moles of S, will need ( 0.00779 .16 ) = 0.124 moles of Ag

(I only have 0.0185 moles of Ag)

So the Ag is the limiting reactant, now we can calculate the mass of formed product:

16 moles of Ag, produce 8 moles of Ag₂S

0.0185 moles of Ag will produce (0.0185 .8)/ 16 = 0.00925 moles of Ag₂S

To find out the mass, let's multiply moles . molar mass

0.00925 m . 247.8 g/m = 2.29 g

Mass of the excess, which is left unreacted:

0.00779 m - 0.00116m = 0.00667 moles

0.00667m . 256.48 g/m = 1.70 g