Answer:
81.08g of H[tex]_{2}[/tex]O will be produced.
Explanation:
Write down the balanced chemical equation:
[tex]B_5H_9 + O_2[/tex] ⇒ [tex]B_2O_3+H_2O[/tex]
[tex]2B_5H_9+12O_2[/tex] ⇒ [tex]5B_2O_3+9H_2O[/tex]
Determine the limiting reagent:
[tex]B_5H_9[/tex] :- 126/63.12646 = 1.995993 mol
1.995993/2 = 0.9979965
[tex]O_2[/tex] :- 192/31.9988 = 6.000225 mol
6.000225/12 = 0.50001875
Therefore, [tex]O_2[/tex], is the limiting reagent.
Use stoichiometry ratios to determine the number of moles of water produced:
[tex]O_2[/tex] : [tex]H_2O[/tex]
12 : 9
6.000225 : 4,500168756328362
Use the mole formula to calculate the mass of water produced:
[tex]n=\frac{m}{M} \\m=nM\\m=(4.500168756328362)(18.01528)\\m=81.08g[/tex]
