Question

# Diborane (B2H6) reacts with water to form boric acid (H3BO3) and hydrogen gas (H2) according to...

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)

a) According to stochiometry of reaction, For every 6 moles of H2​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 .

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