Sulfur and fluorine react to form sulfur hexafluoride: S (s) + 3 F_2 (g) rightar

Question

Sulfur and fluorine react to form sulfur hexafluoride: S (s) + 3 F_2 (g) rightarrow SF_6 (g) If 50.0 g S is allowed to react as completely as possible with 105.0 g F_2 (g), what mass of the excess reactant is left? 20.5 g S 45.7 g F_2 15.0g S 36.3 g F_2 Sulfur and fluorine react to form sulfur hexafluoride: S (s) + 3 F_2 (g) rightarrow SF_6 (g) If 50.0 g S is allowed to react as completely as possible with 105.0 g F_2 (g), what mass of the excess reactant is left? 20.5 g S 45.7 g F_2 15.0 g S 36.3 g F_2

Explanation / Answer

First, balance reaction

S + 3F2 = SF6

then

calculate moles of each species

mol of S = mass/MW = 50/32.065 = 1.5593 moles of S

mol of F2 = mass/MW = 105/37.9968 = 2.76339 moles of F2

clearly, S is in exces, since

2.76339 mol of F2 --> 2.76339/3 = 0.92113 moles of S required

so....

100% reaction implies

0.92113 mol of S react -->

1.5593 - 0.92113 = 0.63817 moles of S left

mass = mol*MW = 0.63817*32 = 20.42144 g

nearest answer is A

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