Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Stoichiometry - Mg & Zn(NO3)2?
I've looked this over and my answer just doesn't seem right...
Here is my chemical equation:
Zn(NO3)2 (aq) + Mg (s) → Mg(NO3)2 (aq) + Zn (s)
I want to know how many mL of zinc nitrate I will need to react completely with 4g of magnesium. I keep getting around 820mL - but that seems way farfetched for a mere 4g! Any help would be appreciated. Please show your process if you can. Here is my presumably incorrect calculation:
(4.0g Mg)/1 ∙ (1 mol Mg)/(24.31g Mg) ∙ (1 mol Zn (NO3)2)/(1 mol Mg) ∙ (1 L)/(0.20 mol Zn(NO3)2 ) ∙ (1000 mL)/(1 L) ≈ 820 mL Zn(NO3)2
Thanks a lot!
1 Answer
- 1 decade agoFavorite Answer
Your answer is correct.
No. moles of Mg = 4/24.3 = 0.165 mol = No. moles of Zn(NO3)2 required
Assuming the concentration of Zn(NO3)2 is 0.20M
Volume = 0.165/0.20 = 822 mL
It does make sense if you think about it. Magnesium is a relatively light metal and so 4g of Magnesium actually contains quite a lot of Mg atoms. Furthermore, you are using a rather dilute solution of Zn(NO3)2, and so you should need a large amount of solution.
