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Answer:
6.499 g
Explanation:
One part of the aqueous HBr reacted with Zinc Hydroxide following the reaction:
- Zn(OH)₂ + 2HBr → ZnBr₂ + 2H₂O
And the remaining HBr reacted with NaOH:
- NaOH + HBr → NaBr + H₂O
First we <u>calculate how many HBr moles remained after reacting with Zn(OH)₂</u>. That number equals the number of NaOH moles used in the titration:
- 0.500 M * 88.5 mL = 44.25 mmol NaOH = mmol HBr
Now we <u>calculate how many moles of HBr reacted with Zn(OH)₂</u>:
- Originally there were (350 mL * 0.500 M) 175 mmol HBr
- 175 mmol - 44.25 mmol = 130.75 mmol HBr
Then we <u>convert those 130.75 mmoles of HBr to the Zn(OH)₂ moles they reacted with</u>:
- 130.75 mmol HBr *
= 65.375 mmol Zn(OH)₂.
Finally we<u> convert Zn(OH)₂ moles to grams</u>:
- 65.375 mmol Zn(OH)₂ * 99.424 mg/mmol = 6499.8 mg Zn(OH)₂
- 6499.8 mg Zn(OH)₂ / 1000 = 6.499 g