Answer:
11.0845g of is produced from the combustion of 96.9 g of
Explanation:
Given:
mass of used in combustion= 96.9 g
To find:
amount of produced=?
Solution:
Let us find the molecular mass of
Atomic weight of Boron(B)=10.811
Atomic weight of Hydrogen(H)=1.007
Now the molecular mass of
=>2(10.811)+6(1.007)
=>21.622 x 6.042
=>27.664
In the given 96.9 g of , the number of moles of
present is
So in 96.9 g of B2H6, 3.50 moles of B2H6 is present which is converted into 2 molecules of
The molecular mass of 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 to HBO2
=>
=>
=>
=>
=>11.0845