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Calculate the mass of water producd from the reaction of 126g of pentaboron noahydride with 192g of moleculAR OXYGEN

Sagot :

shimanekgare

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]

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