Question

What volume, in mL, of a 0.539 M solution of NaBH4 is required to produce 0.579 g of B2H6? H2SO4 is present in excess.

Answer 1

Answer:

70.872 mL

Explanation:

The reaction of Sodium borohydrate with sulfuric acid to form diborane is shown below as:

2 NaBH₄ + 2 H₂SO₄ = B₂H₆ + 3 H₂ + 2 NaSO₄

Given mass of B₂H₆ = 0.579 g

Molar mass of B₂H₆ = 27.66 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles of B₂H₆are:

$moles= \frac{0.579\ g}{27.66\ g/mol}$

$moles= 0.0191\ mol$

From the reaction,

1 mole of B₂H₆ is produced when 2 moles of NaBH₄ react with acid.

So,

0.0191 moles of B₂H₆ is produced when 2×0.0191 moles of NaBH₄ react with acid.

Moles of NaBH₄ required = 0.0382 moles

Given, Molarity of NaBH₄ = 0.539 M

Molarity =  Moles / Volume

Volume = Moles / Molarity = 0.0382 moles / 0.539 M = 0.070872 L

Also, 1L = 1000 mL

So,

Volume = 70.872 mL