Question

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 296.1 g of B2H6?

Answer 1

Answer:

1027.62 g

Explanation:

For $B_2H_6$  :-

Mass of $B_2H_6$  = 296.1 g

Molar mass of $B_2H_6$  = 27.66 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{296.1\ g}{27.66\ g/mol}$

$Moles\ of\ B_2H_6= 10.705\ mol$

From the balanced reaction:-

$B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}$

1 mole of $B_2H_6$ react with 3 moles of oxygen

Thus,

10.705 mole of $B_2H_6$ react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g