Theoretical Yield of SF₆ = 2.01 moles
Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.
Solution 1:
Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.
Given S₈ + 24F₂ => 8SF₆
425g 229g ?
= 425g/256g/mol. = 226g/38g/mol.
= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant
Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant.
S₈ = 1.66/1 = 1.66
F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant
Determining Theoretical Yield:
Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.
Given S₈ + 24F₂ => 8SF₆
425g 229g ?
= 425g/256g/mol. = 226g/38g/mol.
= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant
Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈
Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆ to make the outcome smaller than 6.03. That is,
moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)
S₈ + 24F₂ => 8SF₆
moles SF₆ = 8/24(6.03) moles = 2.01 moles
You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.
This analysis works for all reaction stoichiometry problems.
Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.
Here's another example just for grins ...
C₂H₆O + 3O₂ => 2CO₂ + 3H₂O
Given: 253g 307g ? ?
a. Determine Limiting Reactant
b. Determine mass in grams of CO₂ & H₂O produced
Limiting Reactant
moles C₂H₆O = 253g/46g/mol = 5.5 moles => 5.5/1 = 5.5
moles O₂ = 307g/32g/mol = 9.6 moles => 9.6/24 = 0.4 ∴ O₂ is L.R.
But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.
C₂H₆O + 3O₂ => 2CO₂ + 3H₂O
9.6 mole (L.R.) ? ?
mole yield CO₂ = 2/3(9.6)mole = 6.4 mole (CO₂ coefficient < O₂ coefficient)
mole yield H₂O = 9.6mole = 9.6mole (coefficients O₂ & CO₂ are same.)
mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient C₂H₆O < coefficient O₂)
For grams => moles x formula weight (g/mole)
