Question

Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S (s) 3F 2(g) SF 6 (g) The maximum amount of SF 6 that can be produced from the reaction of 3.5 g of sulfur with of fluorine is ________ g.

Answer 1

Answer:

15.95 g

Explanation:

Calculation of the moles of sulfur as:-

Mass = 3.5 g

Molar mass of sulfur = 32.065 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{3.5\ g}{32.065\ g/mol}$

$Moles= 0.1092\ mol$

From the reaction,

$S+3F_2\rightarrow SF_6$

1 mole of sulfur on reaction forms 1 mole of sulfur hexafluoride

0.1092 mole of sulfur on reaction forms 0.1092 mole of sulfur hexafluoride

Molar mass of sulfur hexafluoride = 146.06 g/mol

Mass= Moles*Molar mass = 0.1092*146.06 g = 15.95 g

15.95 g is the maximum amount of $SF_6$ that can be produced from the reaction of 3.5 g of sulfur with of fluorine.