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.

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:

Thus, moles of B₂H₆are:

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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g