MCQ
Metal carbides on reaction with $\mathrm{H}_2 \mathrm{O}$ forms $\ce{CH_4}$.Carbide can be :
- A$\mathrm{CaC}_2$
- B$\mathrm{Mg}_3 \mathrm{C}_2$
- ✓$\mathrm{Be}_2 \mathrm{C}$
- D$\ce{SiC}$
✓
Answer
Correct option: C.
$\mathrm{Be}_2 \mathrm{C}$
Different metal carbides react with $\mathrm{H}_2 \mathrm{O}$, give the following compounds.
$ \mathrm{Be}_2 \mathrm{C}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{Be}(\mathrm{OH})_2(\mathrm{~s})+\mathrm{CH}_4(\mathrm{~g})$
$ \mathrm{CaC}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_2 \mathrm{H}_2+\mathrm{Ca}(\mathrm{OH})_2 $
$ \mathrm{Mg}_3 \mathrm{C}_2+4 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Mg}(\mathrm{OH})_2+\mathrm{C}_3 \mathrm{H}_4$
Here, beryllium carbide is a metal carbide that forms methane.
$ \mathrm{Be}_2 \mathrm{C}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{Be}(\mathrm{OH})_2(\mathrm{~s})+\mathrm{CH}_4(\mathrm{~g})$
$ \mathrm{CaC}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_2 \mathrm{H}_2+\mathrm{Ca}(\mathrm{OH})_2 $
$ \mathrm{Mg}_3 \mathrm{C}_2+4 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Mg}(\mathrm{OH})_2+\mathrm{C}_3 \mathrm{H}_4$
Here, beryllium carbide is a metal carbide that forms methane.
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