V = 60.0 g/ 0.70 g/mL = 85.7 mL Hope this helps! ;D
The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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I hope it helps you!
First, you need to find the mass of 1 mol of sugar. Mass, or molar mass, can simply be found by adding the masses of the individual elements. These are given to you on the periodic table.

12 x 12.011 grams (molar mass of Carbon) = 144.132 g
22 x 1.008 grams (molar mass of Hydrogen) = 22.176 g
11 x 15.999 grams (molar mass of Oxygen) = 175.989 g
Add all of the pieces together.
144.132 g + 22.176 g + 175.989 g = 342.297 grams
So, if one mole has 342.297 grams, then 7.35 of that amount will be your answer.
342.297 g/mol x 7.35 mol = 2,515.88 grams