Answer :
Answer:
The reaction of 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex] can produce 81 g of Water.
Explanation:
We have the reaction:
[tex]2B_5H_9 + 12O_2 -> 5B_2O_3 + 9H_2O[/tex]
There are 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex].
First step to calculate the water produced is to find the limit reagent. Molecular weights for those substances involved in the chemical reaction are:
[tex]B_5H_9[/tex] = 63.12 g/mol
[tex]O_2[/tex] = 32 g/mol
[tex]B_5O_3[/tex] = 69.62 g/mol
[tex]H_2O[/tex] = 18 g/mol
Now, we can express 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex] as mol dividing the mass using the molecular weight.
[tex]192 g O_2 (\frac{1mol O_2}{32 g O_2} )= 6 mol O2\\ 126 g B_5H_9 (\frac{1mol B_5H_9}{63.12 g B_5H_9} ) = 1.99 mol B_5H_9[/tex]
After that, we should divide the mol of reagent with their respective stoichiometric coefficient to find the limit reagent so:
[tex]6 mol O_2 / 12 mol O_2 = 0.5 \\1.99 mol B_5H_9 /2 mol B_5H_9 = 0.99 \\[/tex]
It means the limit reagent is Oxygen - [tex]O_2[/tex].
Second step to calculate water produced is to use stoichiometric calculations using oxygen amount. According to the balanced equation 12 mols of [tex]O_2[/tex] will produce 9 mols of [tex]H_2O[/tex] . After that, using molecular weight of water [tex]H_2O[/tex] = 18 g/mol we can calculate the mass. It is shown in the next equations:
[tex]6 molO_2 (\frac{9molH_2O}{12molO_2} ) (\frac{18 gH_2O}{1molH_2O} ) = 81 g H_2O[/tex].
Finally, we found that the reaction of 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex] can produce 81 g of Water.