When 10.0 g of lead are heated with 1.6 g of sulfur, 11.6 g of lead sulfide are formed. how many grams of lead sulfide form when 10.0 g of lead are heated with 3.0 g of sulfur?

11.6 g of lead sulfide. First, get the molar masses of lead and sulfur Lead = 207.2 Sulfur = 32.065 Now determine how many moles of each we have avaiable lead = 10.0 g / 207.2 g/mol = 0.048262548 mol sulfur = 1.6 g / 32.065 g/mol = 0.049898643 = mol This tells me that the what's being produced is PbS instead of PbS2 and that there's a very slight excess of sulfur in the original reaction. So on the 2nd reaction with the same amount of lead and twice the amount of sulfur, there will be an even greater excess of sulfur and that you'll get 11.6 g of lead sulfide.