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I am a third year student in junior high chool and I am having some difficulty finding information about the reaction of metal oxides and acids, and I cannot find a more accurate one online. What exactly is produced by the reaction between curpic oxide and nitric acid? Why does the concentration of nitric acid cause the products to be different? Here're something I've found: $\text{CuO}+2\text{HNO}_3\text{(dilute)}=\text{Cu(NO}_3\text{)}_2+\text{H}_2\text{O}$

$\text{CuO}+6\text{HNO}_3\text{(concentrated)}=\text{Cu(NO}_3\text{)}_2+4\text{NO}_2\uparrow+\text{O}_2\uparrow+3\text{H}_2\text{O}$

$2\text{CuO}+8\text{HNO}_3\text{(concentrated)}=2\text{Cu(NO}_3\text{)}_2+4\text{NO}_2\uparrow+\text{O}_2\uparrow+4\text{H}_2\text{O}$

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All your equations are the sums of the two following equations $$\ce{CuO + 2 HNO3 -> Cu(NO3)2 + H2O}$$ and $$\ce{4 HNO3 -> 4 NO2 + O2 + 2 H2O}$$. They can be added with any numerical parameter before being added.

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  • $\begingroup$ Thanks! Btw, do similar ecomposition reactions occur on $\text{HCl}$ and $\text{H}_2\text{SO}_4$? $\endgroup$ Dec 21, 2021 at 10:38
  • $\begingroup$ No ! $\ce{H2SO4}$ does not react with $\ce{HCl}$, because in the $19$th century $\ce{H2SO4}$ was used for synthesizing $\ce{HCl}$, at $300$°C, by the reaction : $\ce{H2SO4(l) + NaCl(s) -> Na2SO4(s) + 2 HCl(g)}$ $\endgroup$
    – Maurice
    Dec 21, 2021 at 13:31
  • $\begingroup$ I think you may have misunderstood me a little (Pardon me, I am not a native speaker of English). I meant that do similar reactions occur when curpic oxide reacts with $HCl$ and $H_2SO_4$ under different concentrations? $\endgroup$ Dec 25, 2021 at 2:00
  • $\begingroup$ @DrThomasPeng. I am like you. I am not a native speaker of English. I will try to answer your question. Cupric oxide reacts with all acids, and in particular with $\ce{HCl}$ and $\ce{H2SO4}$, according to : $$\ce{CuO + 2 HCl -> CuCl2 + H2O}$$ $$\ce{CuO + H2SO4 -> CuSO4 + H2O}$$ and it does not depend on the concentration. The reaction is the same at all concentrations of the acids $\endgroup$
    – Maurice
    Dec 25, 2021 at 11:57
  • $\begingroup$ You have my gratitude. That's a very clear explanation. $\endgroup$ Dec 25, 2021 at 12:24

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