chemistry question

Question

The compound diborane (B2H6) was at one time considered for use as a rocket fuel. Its combustion reaction is B2H6(g) + 3 O2(ℓ) → 2 HBO2(g) + 2 H2O(ℓ) The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation of the investigation of the diborane as a fuel. What mass of liquid oxygen (LOX) would be needed to burn 279.2 g of B2H6? Answer in units of g

Answer 1

Start with the balanced equation:

B₂H₆(g) + 3 O₂(ℓ) → 2 HBO₂(g) + 2 H₂O(ℓ)\

From this balanced equation we see that it takes 3 moles of O₂ to react with 1 mole of B₂H₆

So we will first find out how many moles of B2H6 we have in 279.2 g and then the moles of O2 needed will be 3 x that.

Moles of B₂H₆ present:

molar mass B2H6 = (2x10.81 + 6x1.01) = 21.62 + 6.06 = 27.68 g / mol

moles B₂H₆ = 279.2 g x 1 mol / 27.68 g = 10.09 moles B₂H₆

moles O₂ needed = 10.09 moles B₂H₆ x 3 moles O₂ / 1 mol B₂H₆ = 30.27 moles O₂ needed

Now, simply use the molar mass of O₂ to convert this to grams:

30.27 moles O₂ x 32 g / mol = 968.6 g O₂ needed