Final answer:
The mass of silver sulfide produced from the reaction of 7.2g Ag and 8.0g S is 8.273 g, with Ag being the limiting reactant.
Step-by-step explanation:
To determine the mass of silver sulfide (Ag2S) produced from a mixture of 7.2g of silver (Ag) and 8.0g of sulfur (S), we first need to calculate the moles of each reactant. We then can identify the limiting reactant based on the stoichiometry of the balanced reaction: 2Ag (s) + S(s) → Ag2S (s).
We use the atomic masses to convert the given masses to moles:
1 mol of Ag = 107.87 g, so 7.2 g Ag ÷ 107.87 g/mol = 0.06676 mol Ag
1 mol of S = 32.07 g, so 8.0 g S ÷ 32.07 g/mol = 0.24954 mol S
According to the reaction, each mole of S reacts with 2 moles of Ag. Therefore, for 0.24954 mol of S, we would need 0.49908 mol of Ag. Since we only have 0.06676 mol of Ag, Ag is the limiting reactant.
Next, we calculate the moles of Ag2S produced by doubling the moles of Ag, since 2 moles of Ag produce 1 mole of Ag2S: 0.06676 mol Ag → 0.03338 mol Ag2S.
Finally, we convert moles of Ag2S to mass:
1 mol of Ag2S = 247.8 g (107.87 × 2 + 32.07 g), so 0.03338 mol Ag2S × 247.8 g/mol = 8.273 g of Ag2S.
Therefore, 8.273 g of silver sulfide will be produced from the reaction of 7.2g Ag and 8.0g S.