Answer:
11.0845g of [tex]HBO_2[/tex] is produced from the combustion of 96.9 g of [tex]B_2H_6[/tex]
Explanation:
Given:
mass of [tex]B_2H_6[/tex] used in combustion= 96.9 g
To find:
amount of [tex]HBO_2[/tex] produced=?
Solution:
Let us find the molecular mass of[tex]B_2H_6[/tex]
Atomic weight of Boron(B)=10.811
Atomic weight of Hydrogen(H)=1.007
Now the molecular mass of [tex]B_2H_6[/tex]
=>2(10.811)+6(1.007)
=>21.622 x 6.042
=>27.664
In the given 96.9 g of [tex]B_2H_6[/tex], the number of moles of [tex]B_2H_6[/tex]present is
[tex]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[/tex]
So in 96.9 g of B2H6, 3.50 moles of B2H6 is present which is converted into 2 molecules of [tex]HBO_2[/tex]
The molecular mass of [tex]HBO_2[/tex] 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 [tex]B_2H_6[/tex] to HBO2 [tex]HBO_2[/tex]
=>[tex]3.50 \text { moles of } B_2 H _6 \times \frac{2 \text { moles of } HBO_2}{1 \text { mole of } B _2H _6}}[/tex]
=>[tex]3.50\times \frac {2\times (43.816)}{27.666}[/tex]
=>[tex]3.50\times \frac{87.632}{27.666}[/tex]
=>[tex]3.50\times 3.167[/tex]
=>11.0845
