How many moles of zinc, Zn, are in 1.31×1024 Zn atoms?
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Answer:
0.0457 M
Explanation:
The reaction that takes place is:
- 2HBr + Ca(OH)₂ → CaBr₂ + 2H₂O
First we<u> calculate how many moles of acid reacted</u>, using the <em>HBr solution's concentration and volume</em>:
- Molarity = Moles / Volume
- Molarity * Volume = Moles
- 0.112 M * 12.4 mL = 1.389 mmol HBr
Now we <u>convert HBr moles to Ca(OH)₂ moles</u>, using the stoichiometric ratio:
- 1.389 mmol HBr *
= 0.6944 mmol Ca(OH)₂
Finally we <u>calculate the molarity of the Ca(OH)₂ solution</u>, using the <em>given volume and calculated moles</em>:
- 0.6944 mmol Ca(OH)₂ / 15.2 mL = 0.0457 M