There is 6g of (+) rotation species and 4g of (-) rotation species. The 4g of (-) rotation species will cancel the effect of 4g of (+) rotation species. So there will be an excess of 6-4= 2g (+) rotation species left.
Enantiomeric excess= (Excess species)/(Total sample) x 100 = [2/(6+4)] x 100 = 20%

To calculate the enantiomeric excess (ee) of a mixture, we need to determine the difference in the amounts of the two enantiomers present. In this case, we have a mixture containing 6g of (+) 2-butanol and 4g of (-) 2-butanol.

Step 1: Determine the molar masses of the enantiomers
The molar mass of 2-butanol is 74.12 g/mol.

Step 2: Calculate the moles of each enantiomer
Moles = Mass / Molar mass

For (+) 2-butanol:
Moles of (+) 2-butanol = 6g / 74.12 g/mol

For (-) 2-butanol:
Moles of (-) 2-butanol = 4g / 74.12 g/mol

Step 3: Calculate the total moles of both enantiomers
Total moles = Moles of (+) 2-butanol + Moles of (-) 2-butanol

Step 4: Calculate the enantiomeric excess
Enantiomeric excess = (Moles of (+) 2-butanol - Moles of (-) 2-butanol) / Total moles * 100

Let's now plug in the values and calculate the enantiomeric excess:

Moles of (+) 2-butanol = 6g / 74.12 g/mol = 0.081 mol
Moles of (-) 2-butanol = 4g / 74.12 g/mol = 0.054 mol
Total moles = 0.081 mol + 0.054 mol = 0.135 mol

Enantiomeric excess = (0.081 mol - 0.054 mol) / 0.135 mol * 100 = 20%

Therefore, the enantiomeric excess of the mixture is 20%, which corresponds to option b.

In conclusion, the enantiomeric excess of a mixture containing 6g of (+) 2-butanol and 4g of (-) 2-butanol is 20%. This means that there is a 20% excess of one enantiomer over the other in the mixture.