Answer:
Mole ratio of o-phen to Fe²⁺ is 6.
Explanation:
In the experiment, you are using 0.00136M Fe²⁺ and 0.00407M o-phen solutions. A stoichiometric volume ratio of 2.0mL o-phen to 1.0mL Fe²⁺ means you need to add 2.0mL of o-phen solution for a complete reaction of 1.0mL of Fe²⁺ solutions.
Moles in both solutions are:
Fe²⁺: 1.0x10⁻³L Fe²⁺ × (0.00136mol Fe²⁺ / 1L) = 1.36x10⁻⁶mol Fe²⁺
o-phen: 2.0x10⁻³L o-phen × (0.00407mol o-phen / 1L) = 8.14x10⁻⁶mol o-phen
That means 8.14x10⁻⁶mol o-phen reacts per 1.36x10⁻⁶mol Fe²⁺. The mole ratio is:
8.14x10⁻⁶mol o-phen / 1.36x10⁻⁶mol Fe²⁺ = 6
Mole ratio of o-phen to Fe²⁺ is 6.
Answer: The stoichiometric mole ratio of o-phen to Fe2+ is 6.0
Explanation: From the question,
Concentration of Fe2+ = 0.00136 M
Concentration of o-phen = 0.00407 M
Volume of Fe2+ = 1.0 mL
Volume of o-phen = 2.0 mL
To determine the stoichiometric mole ratio of o-phen to Fe2+, we will determine the number of moles of each entities that reacted separately.
Using the formula,
Number of moles (n) = Concentration (C) × Volume (V)
For Fe2+
Number of moles (n) = 0.00136 × 1.0
n = 0.00136 moles of Fe2+
For o-phen
Number of moles (n) = 0.00407 × 2.0
n = 0.00814 moles of o-phen
Now, the stoichiometric mole ratio of o-phen to Fe2+ is by
Number of moles of o-phen ÷ Number of moles of Fe2+
= 0.00814 ÷ 0.00136
= 5.985294118
= 6.0 (to two significant figures)
Hence, the stoichiometric mole ratio of o-phen to Fe2+ is 6.0
