Question

Diborane (B2H6) reacts with water to form boric acid (H3BO3) and hydrogen gas (H2) according to the following reaction: B2H6(g) + 6H20(L)----> 2H3BO3(s)+ 6H2(g)

[Molar Masses (g/mol): B2H6:27.67, H2O: 18.02, H3BO3: 61.83, H2:2.02]

a.) In a particular experiment 18.0 mol of H2O are consumed, how many mol of H3BO3 were produced?

b.) Calculate the mass (in g) of B2H6 needed to produce 278.2g of H3BO3 (assuming excess H2O)

Answer 1

a) According to stochiometry of reaction, For every 6 moles of H_2​O consumed 2 moles of H₃BO₃ is produced.

18 moles of H₂O is equal to 3 * 6 = 18. Therefore 3 * 2 = 6 molecules of H₃BO₃ is produced .

b) Number of moles of H₃BO₃ in 278.2 g = weight/ molecular weight= 278.2/61.83= 4.5 moles.

For every 1 mole of B₂H₆ consumed, 2 moles of H₃BO₃ is produced. Therefore for 4.5 moles of H₃BO₃ produced , 4.5/2 = 2.25 moles of B₂H₆ is consumed, which is equivalent to 2.25 * 27.67 ( molecular weight) =62.26 g .