Question

Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S(g) + 3F2(g) ->SF6(g) If 50 g S is allowed to react as completely as possible with 105.0g F2(g), how many grams ofn SF6 can we get?

Answer 1

The quantity of SF6 that we can get is = 228.2g

Calculation of product formation

The quantity of S that reacted = 50g

The quantity of F2(g)that reacted = 105.0g

Therefore the quantity of SF6 that will be produced = ?

But molar mass of sulphur = 32g/mol

molar mass of SF6(g)= 146.06 g/mol

If 32g of S = 146.06 g of SF6

50 g of S = x

Make X the subject of formula,

X = 50×146.06/32

X = 7303/32

X = 228.2g of S

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