A mixture of methane,ethylene,and acetylene g | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Acetylene $(CH \equiv CH)$ reacts with ammoniacal cuprous chloride to form a red precipitate of copper acetylide $(Cu-C \equiv C-Cu)$.$CH \equiv C

Vedclass · Accepted Answer

$A$ mixture of methane and ethylene

Vedclass · Answer

(C) Acetylene $(CH \equiv CH)$ reacts with ammoniacal cuprous chloride to form a red precipitate of copper acetylide $(Cu-C \equiv C-Cu)$.$CH \equiv CH + 2[Cu(NH_3)_2]OH 	o Cu-C \equiv C-Cu + 4NH_3 + 2H_2O$Methane $(CH_4)$ and ethylene $(C_2H_4)$ do not possess acidic hydrogen atoms and therefore do not react with ammoniacal cuprous chloride.$CH_4 + 	ext{Ammoniacal } Cu_2Cl_2 	o 	ext{No reaction}$$C_2H_4 + 	ext{Ammoniacal } Cu_2Cl_2 	o 	ext{No reaction}$Thus,the gases that pass through the bottle without reacting are methane and ethylene.

$A$ mixture of methane,ethylene,and acetylene gases is passed through a Wolf's bottle containing ammoniacal cuprous chloride. The gas coming out is

Similar Questions

$CH_3-C \equiv C-O-CH_3 \xrightarrow{dil. \ H_2SO_4, HgSO_4} \text{major product}$

$CH \equiv CH \mathop {\xrightarrow{Ni(CN)_2}}\limits_{Pressure} X$. Here,$X$ in the reaction is:

The ozonolysis $(O_3, H_2O)$ of $(CH_3)_2CH-C \equiv C-CH_3$ gives:

$A$ metallic carbide on treatment with water gives a colourless gas which burns readily in air and which gives a precipitate with ammoniacal silver nitrate solution. The gas evolved is:

When $23 \ g$ of $Na$ reacts with $1 \ mole$ of propyne,how many moles of $H_2$ gas will be released?

Vedclass Test Series

Exam Paper Generator

Online Exam Module