Question

Sulfur and fluorine react to form sulfur hexafluoride:

S(s)+3F2(g)→SF6(g)

Part A

If 50.0 g S is allowed to react as completely as possible with 105.0 g F2(g), what mass of the excess reactant is left?

If 50.0   is allowed to react as completely as possible with 105.0  , what mass of the excess reactant is left?

	36.3 g F2
	20.5 gS
	7.5 gF2
	15.0 g S

Answer 1

Answer – We are given , mass of S = 50.0 g , mass of F₂ = 105.0 g

Reaction - S(s)+3F₂(g) -----> SF₆(g)

Calculating the moles of each reactant –

Moles of S = 50.0 g / 32.066 g.mol^-1

                   = 1.56 moles of S

Moles of F₂ = 105.0 g / 37.996 g.mol^-1

                      = 2.76 moles of F₂

Calculating the limiting reactant –

Moles of SF₆(g) from moles of S

From the balanced equation –

1 moles of S = 1 moles of SF₆(g)

So, 1.56 moles of s = ?

= 1.56 moles of SF₆(g)

Moles of SF₆(g) from F₂

From the balanced equation –

3 moles of F₂ = 1 moles of SF₆(g)

So, 2.76 moles of s = ?

= 0.92 moles of SF₆(g)

So moles of SF₆(g) is lowest from the moles of F₂, so limiting reactant is F₂ and moles of SF₆(g) is 0.92 moles

Excess mass of S

3 moles of F₂ = 1 moles of S

So, 2.76 moles of F₂ = ?

= 0.92 moles of S

So moles of S remaining = 1.56 -0.92

                                       = 0.639 moles of S

Mass of S excess remaining after reaction = 0.639 moles * 32.066 g/mol

                                                                   = 20.5 g of S

So answer is 20.5 g of S