Question

The maximum amount of SF6 that can be produced from the reaction of 3.5 g of sulfur with 4.5 g of fluorine is __________ g.

Answer 1

16.06 g

Step-by-step explanation:

Reaction equation;

S(s) + 3F2(g) ----> SF6(s)

Number of moles of sulphur reacted = 3.5g/32g/mol = 0.11 moles

If 1 mole of sulphur yields 1 mole of SF6

Then 0.11 moles of sulphur yields 0.11 moles of SF6

Also

Number of moles of fluorine = 4.5g/38 g/mol= 0.12 moles

Hence sulphur his the limiting reactant;

Maximum amount of SF6 = 0.11 × 146 = 16.06 g