The redox reaction is

Here
Calcium undergoes reduction, and acts as cathode
Lithium undergoes oxidation and acts as anode
The reduction potential of calcium is -2.87 V
The reduction potential of lithium is - -3.05 V
We know that
Ecell = Ecathode - Eanode
Ecell = -2.87 - (-3.05) = 0.18 V
Answer is "0.05 mol".
<em>Explanation;</em>
We can do calculation by using a simple formula as
n = m/M
Where, n is the number of moles of the substance (mol), m is the mass of the substance (g) and M is the molar mass of the substance (g/mol).
Here,
n = ?
m = 2.80 g
M = 56.08 g/mol
By substitution,
n = 2.80 g /56.08 g/mol
n = 0.0499 mol ≈ 0.05 mol
Suspension. The particles are big enough for the eye to see, and will separate if left sitting.
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