Question

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

Answer 1

81.08g of H
`_(2)`O will be produced.

Step-by-step explanation:

Write down the balanced chemical equation:

`B_5H_9 + O_2` ⇒
`B_2O_3+H_2O`

`2B_5H_9+12O_2` ⇒
`5B_2O_3+9H_2O`

Determine the limiting reagent:

`B_5H_9` :- 126/63.12646 = 1.995993 mol

1.995993/2 = 0.9979965

`O_2` :- 192/31.9988 = 6.000225 mol

6.000225/12 = 0.50001875

Therefore,
`O_2`, is the limiting reagent.

Use stoichiometry ratios to determine the number of moles of water produced:

`O_2` :
`H_2O`

12 : 9

6.000225 : 4,500168756328362

Use the mole formula to calculate the mass of water produced:

`n=(m)/(M) \\m=nM\\m=(4.500168756328362)(18.01528)\\m=81.08g`