How long should I leave a solution of zinc oxide powder and heinz vinegar in order for it to react as completely as possible? The solution is 1(zinc oxide powder):2(heinz vinegar) in a glass flask. How much is this process increased by heat (say 50 deg C)? Without heat, should I cover the flask to prevent the acetic acid from evaporation while it reacts?
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2$\begingroup$ Heinz vinegar is not a chemical substance to determine the ratio. If the vinegar contained 60 g acetic acid/L (about 6%), than you need minimally 2 L of vinegar for about 81 g of zinc oxide, but little excess would be needed. $\endgroup$– PoutnikSep 17, 2022 at 1:42
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1$\begingroup$ So the short answer to the title question is: It would never fully react. $\endgroup$– PoutnikSep 17, 2022 at 2:49
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$\begingroup$ @Poutnik thank you so much for this response! How long would it take for the reaction to occur at room temp vs. gentle heat/stovetop? $\endgroup$– tomatopiesSep 17, 2022 at 17:33
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$\begingroup$ And also just for future reference, what kind of calculation is done to determine how long these kind of reactions take? I now know you can use stoichiometry to figure out the ratios (lol), so what's the kind for duration as a function of temp? $\endgroup$– tomatopiesSep 17, 2022 at 17:39
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$\begingroup$ Heterogenous reactions have too many unknown variables, so nothing beats empirical data. As a rule of thumb, reaction rates increase typically 2-4 times per 10 °C increase. $\endgroup$– PoutnikSep 17, 2022 at 19:14
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You might be far better off to react vinegar with zinc carbonate, yielding zinc acetate and CO2.
Zinc carbonate is much more soluble in water, no need for heating, just stir a bit.
answered Sep 16, 2022 at 21:59
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$\begingroup$ Thanks for this I will definitely do that next time - but zinc oxide is what I have on hand.... $\endgroup$ Sep 16, 2022 at 22:18
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$\begingroup$ @tomatopies you can try it with ZnO. Just leave it overnight. Beware acetic acid fumes if you reflux with heat (do it in a hood). If no luck consider going to ZnCl with HCl to get it into solution, then make the carbonate with baking soda. $\endgroup$ Sep 17, 2022 at 0:46