$$\ce{Mg + C -> X}$$ $$\ce{X + H2O -> Y}$$ Determine unknowns X and Y.

Now the problem is I think the answer should be $\ce{X=MgC2}$ and $ \ce{Y=C2H2 + Mg(OH)2}$. But the given answer was $ \ce{X}= \ce{Mg2C3}$ and $\ce{Y}=\ce{C3H4} + \ce{Mg(OH)2}$

I tried to google the answer but where some website say $\ce{C3H4}$ is correct some say $\ce{C2H2}$ is. Funny thing is in the exam just after that one we got the same question and this time the $\ce{C2H2}$ was the answer (and I wrote $\ce{C3H4}$)

  • 5
    $\begingroup$ Magnesium forms more than one carbide. The question is ill posed. $\endgroup$ – Oscar Lanzi Mar 3 '18 at 0:25
  • 3
    $\begingroup$ @OP: Were there any other conditions in the question? Or was the question literally word-for-word your equations? $\endgroup$ – JavaScriptCoder Mar 3 '18 at 1:20
  • $\begingroup$ Yes, both type of carbide is possible. At a certain condition, one type of carbide predominates over the other. $\endgroup$ – Nilay Ghosh Mar 3 '18 at 6:00
  • $\begingroup$ @JavaScriptCoder The equation arrow did have a delta sign over it indicating that it's heated but I think it won't help? $\endgroup$ – SmarthBansal Mar 3 '18 at 10:09

This asnswer is intended to extend my comment. Most of the information is taken from this site.

Both types of magnesium carbide($\ce{MgC2, Mg2C3}$) is formed when acetylene gas is passed over magnesium. At different temperatures, one type of carbide predominates over the other. At 570°C-610° C, $\ce{MgC2}$ converts to $\ce{Mg2C3}$:

$$\ce{2MgC2 -> Mg2C3 + C }$$

Above 610° C, $\ce{Mg2C3}$ decomposes:

$$\ce{Mg2C3 -> 2Mg + C}$$

Both types of magnesium carbide can be hydrolysed:

$$\ce{MgC2 + 2H2O -> Mg(OH)2 + C2H2}$$ $$\ce{Mg2C3 + 4H2O -> 2Mg(OH)2 + C3H4}$$

For more information on the structure of magnesium carbide, see this old paper from 1943. Also, how magnesium carbide behaves at high pressure is discussed in this new paper(2015). Another type of magnesium carbide, $\ce{Mg2C}$ exist at high pressure and it along with $\ce{MgC2}$ behaves as semiconductor while $\ce{Mg2C3}$ behave as insulator-metal transition. (OP is requested to see both papers if interested)

  • $\begingroup$ Just summarizing; So X would by MgC2 and then if enough heat is provided it would convert to Mg2C3. Then Y will depend on the temperature at which the reaction is carried out, right? $\endgroup$ – SmarthBansal Mar 3 '18 at 10:07
  • $\begingroup$ Yes, that's the point. $\endgroup$ – Nilay Ghosh Mar 3 '18 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.