Question

Given this unbalanced reaction, how many grams of Boric acid (H3BO3) will you have if you react 25.0 grams of diboron hexahydride with excess water? Follow Significant Figures

B2H6 + H2O -> H3BO3 + H2

Answer 1

The balanced reaction is as following:

B₂H₆ + 6H₂O -> 2H₃BO₃ + 6H₂

The mass of boric acid produced can be calculated as:

m(H₃BO₃) = m(B₂H₆) × 2 × Mr(H₃BO₃) / Mr(B₂H₆).

As Mr(B₂H₆) = 27.66 g/mol and Mr(H₃BO₃) = 61.83 g/mol:

m(H₃BO₃) = 25.0 g × 2 × 61.83 g/mol / 27.66 g/mol = 111.8 g

Answer: 111.8 g