To determine the minimum amount of BH3 required for the complete hydroboration-oxidation of 105g of 1-pentene, we need to calculate the stoichiometry of the reaction.

1. Balanced Chemical Equation:
The hydroboration-oxidation reaction of 1-pentene with BH3 can be represented by the following balanced chemical equation:
C5H10 + BH3 → C5H11B + H2O

2. Molecular Mass Calculation:
The molecular mass of 1-pentene (C5H10) can be calculated as follows:
(5 x 12.01 g/mol) + (10 x 1.01 g/mol) = 70.15 g/mol

3. Stoichiometry Calculation:
By comparing the balanced chemical equation, we can observe that one mole of 1-pentene reacts with one mole of BH3 to produce one mole of C5H11B. Therefore, the molar mass of 1-pentene is equal to the molar mass of C5H11B.

4. Moles of 1-Pentene:
To determine the moles of 1-pentene present in 105g, we divide the given mass by the molar mass of 1-pentene:
Moles of 1-pentene = 105g / 70.15 g/mol = 1.497 mol

5. Moles of BH3:
Since the stoichiometry of the reaction is 1:1, the moles of BH3 required will also be the same as the moles of 1-pentene.

6. Mass of BH3:
To calculate the mass of BH3 required, we multiply the moles of BH3 by its molar mass:
Mass of BH3 = Moles of BH3 x Molar mass of BH3
Mass of BH3 = 1.497 mol x 14.01 g/mol ≈ 20.96 g

Therefore, the minimum amount of BH3 required for the complete hydroboration-oxidation of 105g of 1-pentene is approximately 20.96 grams.

However, the given correct answer is '7' grams. This suggests that there may be additional information or specific conditions provided in the question that require a different approach or consideration.