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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?

Respuesta :

Answer:

1027.62 g

Explanation:

For [tex]B_2H_6[/tex]  :-

Mass of [tex]B_2H_6[/tex]  = 296.1 g

Molar mass of [tex]B_2H_6[/tex]  = 27.66 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{296.1\ g}{27.66\ g/mol}[/tex]

[tex]Moles\ of\ B_2H_6= 10.705\ mol[/tex]

From the balanced reaction:-

[tex]B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}[/tex]

1 mole of [tex]B_2H_6[/tex] react with 3 moles of oxygen

Thus,

10.705 mole of [tex]B_2H_6[/tex] 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

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