Stoichiometry Problem

Question

1. 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 3.0 g Ag and 3.0 g S₈?

                                            16Ag + S₈ → 8Ag₂S

Answer 1

Hello, Emma,

The first thing we need to determine is whether one of the two reactants will limit how far the reaction will proceed (i.e., will one be completely consumed before the other?). Convert grams of the reactants into moles:

3 grams Ag is 3g/107.9 g/mole = 0.0278 moes Ag

3 grams S₈ is 3g/256.5g/mole = 0.0117 moles S₈

The balanced equation says we need 16 moles of Ag for every 1 mole of S₈. Conversely, we need 1/16 mole of S₈ for every mole of Ag. If we take 1/16 of 0.0278 mole Ag, that would mean 0.00087 mole of S₈ is consumed by the time we react all of the Ag. We have far more S₈ than needed, so that means Ag is the limiting reagent. We'll base the product calculations on the assumpotion that all of nthe silver is consumed. It could be noted that the remaining S₈ will be (0.0117 - 0.00087)= 0.0108 moles.

The equation tells us that we'll get 8 moles of Ag₂S for every 16 moles Ag, a molar ratio of 1/2, or 0.5. Since Ag is the limiting reagent, we can assume all of it is consumed, generating (0.5)*(0.02781 moles) = 0.01391 moles of Ag₂S.

Multiply 0.01391 moles of Ag₂S by it's molar mass of 247.8 g/mole to arrive at a mass of 3.445 grams Ag₂S. 2 sig figs would be 3.4 grams, but I find it painful to give up hard-won digits, so I'll leave that to you.

Bob