Question

Consider the reaction:

2 NaBH₄(aq) + H₂SO₄(aq) → 2 H₂(g) + Na₂SO₄(aq) + B₂H₆(g)

What volume, in mL, of a 0.515 M solution of NaBH₄ is required to produce 0.541 g of B₂H₆? H₂SO₄ is present in excess.

mL

Answer 1

2 NaBH₄(aq) + H₂SO₄(aq) → 2 H₂(g) + Na₂SO₄(aq) + B₂H₆(g)

Number of moles of B₂H₆ = amount of B₂H₆ (g)/ mw of B₂H₆ (g/mol)

                                                     = 0.541 g/27.66 g/mol= 0.0195589 moles

According to above balance equation

1 mole of B₂H₆ is produced by 2 moles of NaBH₄

Therefore 0.0195589 moles will be produced by = 2 X 0.0195589 moles of NaBH4= 0.0391178 moles NaBH4

Number of moles of NaBH4 required= Molarity of NaBH4 X volume of NaBH4 in L

0.0391178 mole= 0.515 mol/L X volume of NaBH4 in L

volume of NaBH4 in L= 0.075957 L

                                    = 75.95 mL Answer