Question

The compound diborane (b2h6) was at one time considered for use as a rocket fuel. its combustion reaction is b2h6(g) + 3 o2(â) â 2 hbo2(g) + 2 h2o(â) 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 106.3 g of b2h6?

Answer 1

To burn 106.3 g of diborane (B₂H₆), 368.64 g of liquid oxygen (LOX) would be required. This is calculated based on the stoichiometry of the balanced chemical combustion reaction.

Step-by-step explanation:

To calculate the mass of liquid oxygen (LOX) needed to burn 106.3 g of diborane (B₂H₆), we must first use the balanced chemical equation for the combustion of diborane:

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

From the equation, we can see that 1 mole of B₂H₆ reacts with 3 moles of O₂. First, determine the molar mass of B₂H₆ which is (2 × 10.81) + (6 × 1.008) = 27.7 g/mol. Now, divide the given mass of B₂H₆ by its molar mass to get the number of moles of B₂H₆:

106.3 g B₂H₆ × (1 mol B₂H₆/27.7 g) = 3.84 moles of B₂H₆

Now calculate the moles of O₂ required using the stoichiometry of the reaction (3 moles of O₂ per mole of B₂H₆):

3.84 moles of B₂H₆ × 3 moles O₂/1 mole B₂H₆ = 11.52 moles O₂

Finally, calculate the mass of O₂ using the molar mass of O₂ which is (2 × 16.00) = 32.00 g/mol:

11.52 moles of O₂ × 32.00 g/mol = 368.64 g of O₂

So, 368.64 g of LOX would be needed to burn 106.3 g of diborane (B₂H₆).

Answer 2

The atomic mas of Oxygen gas is 32 g/mol
The molar mass of B2H6 IS 27.66 g/mol
Therefore, a mass of 106.3 g of B2H6 will contain;
106.3/27.66 = 3.843 moles
The mole ratio of B2H6 : O2 is 1: 3
Therefore, moles of oxygen will be 3.843 ×3/1 = 11.529 moles
Therefore, the mass of liquid oxygen needed would be;
11.529 × 32
= 368.94 g