If 40.0 g s is allowed to react as completely as possible with 150.0 g f2( g., what mass of the excess reactant is left?

7.80 g of Fluorine will be left over. First, lookup the atomic weights of sulfur and fluorine Sulfur = 32.065 Fluorine = 18.998403 Now calculate how many moles of each reactant we have Sulfur = 40.0 / 1.247466 = 1.247466 Fluorine = 150.0 / 18.998403 = 7.8954 The produce will be Sulfur hexafluoride which is SF6. So for every mole of sulfur you need 6 moles of Fluorine . 1.247466 * 6 = 7.484796507 Since 7.484796 is less than 7.8954, we have an excess of Fluorine. Let's subtract the amount of fluorine we need from what we have. 7.8954 - 7.4848 = 0.4106 And the mass of 0.4106 moles of Fluorine is 0.4106 mol * 18.998403 g/mol = 7.800744 g Round the result to 3 significant figures.

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.