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.

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