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.
