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), what mass of the excess reactant is left.

Answer 1

The mole number of 50 g S is 1.563 mol. 105.0 g F2 is 2.763 mol. The ratio of S and F2 is 1:3. So the excess one is S. And mass left is 0.642*32=20.52 g.

Answer 2

Answer: The mass of excess reagent left is 20.48 g

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For sulfur:

Given mass of sulfur = 50 g

Molar mass of sulfur = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of sulfur}=(50g)/(32g/mol)=1.56mol`

• For fluorine gas:

Given mass of fluorine gas = 105 g

Molar mass of fluorine gas = 37.99 g/mol

Putting values in equation 1, we get:

`\text{Moles of fluorine gas}=(105g)/(37.99g/mol)=2.76mol`

The chemical equation for the reaction of sulfur and fluorine gas follows:

`S(g)+3F_2(g)\rightarrow SF_6(g)`

By Stoichiometry of the reaction:

3 moles of fluorine gas reacts with 1 mole of sulfur

So, 2.76 moles of fluorine gas will react with =
`(1)/(3)* 2.76=0.92mol` of sulfur

As, given amount of sulfur is more than the required amount. So, it is considered as an excess reagent.

Thus, fluorine gas is considered as a limiting reagent because it limits the formation of product.

Moles of excess reagent (sulfur) left = 1.56 - 0.92 = 0.64 moles

• Now, calculating the mass of sulfur from equation 1, we get:

Molar mass of sulfur = 32 g/mol

Moles of sulfur = 0.64 moles

Putting values in equation 1, we get:

`0.64mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=20.48g`

Hence, the mass of excess reagent left is 20.48 g