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.

The mass of excess reagent left is 20.48 g

Explanation:

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

     .....(1)

Given mass of sulfur = 50 g

Molar mass of sulfur = 32 g/mol

Putting values in equation 1, we get:

Given mass of fluorine gas = 105 g

Molar mass of fluorine gas = 37.99 g/mol

Putting values in equation 1, we get:

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

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 = 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

Molar mass of sulfur = 32 g/mol

Moles of sulfur = 0.64 moles

Putting values in equation 1, we get:

Hence, the mass of excess reagent left is 20.48 g

h let me think