Answer :
Answer:
[tex]\large \boxed{\text{5 mol}}[/tex]
Explanation:
1. Gather all the information in one place.
S + 3F₂ ⟶ SF₆
n/mol: 6 15
They have given us the amounts of two reactants and asked us to calculate the amount of product. This is a limiting reactant problem.
2. Calculate the moles of SF₆ you can obtain from each reactant
From S:
[tex]\text{Moles of SF}_{6} = \text{6 mol S} \times \dfrac{\text{1 mol SF}_{6}}{\text{1 mol S}} = \text{6 mol SF}_{6}[/tex]
From F₂:
[tex]\text{Moles of SF}_{6} = \text{15 mol F}_{2} \times \dfrac{\text{1 mol SF}_{6}}{\text{3 mol F}_{2}} = \text{5 mol SF}_{6}[/tex]
SF₆ is the limiting reactant, because it gives fewer moles ( 5 mol) of SF₆.
The reaction can form 5 mol of SF₆.