Question

What mass of HBO2 is produced from the
combustion of 96.9 g of B2H6?

Answer 1

11.0845g of
`HBO_2` is produced from the combustion of 96.9 g of
`B_2H_6`

Step-by-step explanation:

Given:

mass of
`B_2H_6` used in combustion= 96.9 g

To find:

amount of
`HBO_2` produced=?

Solution:

Let us find the molecular mass of
`B_2H_6`

Atomic weight of Boron(B)=10.811

Atomic weight of Hydrogen(H)=1.007

Now the molecular mass of
`B_2H_6`

=>2(10.811)+6(1.007)

=>21.622 x 6.042

=>27.664

In the given 96.9 g of
`B_2H_6`, the number of moles of
`B_2H_6`present is

`96.9 g \text { of } B_2 H_6 * \frac{1 \text { mol of } B_2 H_6}{27.666g B_2 H _6}=3.50 \text { moles of } B _2 H _6`

So in 96.9 g of B2H6, 3.50 moles of B2H6 is present which is converted into 2 molecules of
`HBO_2`

The molecular mass of
`HBO_2` is

Atomic weight of Boron(B)=10.811

Atomic weight of Hydrogen(H)=1.007

Atomic weight of Oxygen(O)=15.9994

=> 10.811+1.007+2(15.9994)

=>10.811+1.007+31.998

=>43.816

Convert 3.50 moles of
`B_2H_6` to HBO2
`HBO_2`

=>
`3.50 \text { moles of } B_2 H _6 * \frac{2 \text { moles of } HBO_2}{1 \text { mole of } B _2H _6}}`

=>
`3.50* \frac {2* (43.816)}{27.666}`

=>
`3.50* (87.632)/(27.666)`

=>
`3.50* 3.167`

=>11.0845