Question

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

Answer 1

Answer:

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

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

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

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  $HBO_2$

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

=>$3.50\times \frac {2\times (43.816)}{27.666}$

=>$3.50\times \frac{87.632}{27.666}$

=>$3.50\times 3.167$

=>11.0845