Question

Based on the balanced equation below, how many moles of SF6 can form from the combination of 6 moles of S and 15 moles of F2?

Answer 1

`\large \boxed{\text{5 mol}}`

Step-by-step 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:

`\text{Moles of SF}_(6) = \text{6 mol S} * \frac{\text{1 mol SF}_(6)}{\text{1 mol S}} = \text{6 mol SF}_(6)`

From F₂:

`\text{Moles of SF}_(6) = \text{15 mol F}_(2) * \frac{\text{1 mol SF}_(6)}{\text{3 mol F}_(2)} = \text{5 mol SF}_(6)`

SF₆ is the limiting reactant, because it gives fewer moles ( 5 mol) of SF₆.

The reaction can form 5 mol of SF₆.