The concentrations of Fe2+ and o-phen used in an experiment were 0.00136 M and 0.00407 M, respectively. If a student found a stoichimetric volume ratio of 2.0 mL o-phen to 1.0 mL Fe2+, what is the stoichiometric mole ratio of o-phen to Fe2+? Give your answer to two significant figures.

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.

Abdulazeez10 Abdulazeez10 · Answer 2 · 2020-04-18T04:25:07+02:00

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