Question:
1 L of CH\(_4\)(g) and 2 L of Cl\(_2\)(g) react to give HCl(g) and CCl\(_4\)(l) at STP. If 50% of CH\(_4\) is reacted, what will be the total volume of reaction mixture at the same temperature and pressure?
1 L of CH\(_4\)(g) and 2 L of Cl\(_2\)(g) react to give HCl(g) and CCl\(_4\)(l) at STP. If 50% of CH\(_4\) is reacted, what will be the total volume of reaction mixture at the same temperature and pressure?
Show Hint
Liquids formed in reactions are not included in gaseous volume calculations at STP.
Updated On: Jul 18, 2026
- 1.0 L
- 2.5 L
- 3.0 L
- 1.5 L
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understand reaction and physical states.
Methane reacts with chlorine to form carbon tetrachloride and hydrogen chloride. Importantly, CCl\(_4\) is a liquid at STP, so it will not contribute to gaseous volume in the final mixture. This is crucial for volume calculation.
Step 2: Write balanced reaction.
\[ CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl \] This shows stoichiometric relationship between reactants and products in gaseous form.
Step 3: Determine amount of CH\(_4\) reacted.
Given 1 L CH\(_4\), 50% reacts: \[ \text{Reacted CH}_4 = 0.5 \, L \] Unreacted CH\(_4\) = 0.5 L remains in gaseous state.
Step 4: Calculate products formed.
From 0.5 L CH\(_4\): - CCl\(_4\) formed = 0.5 L (liquid, not counted in gas volume) - HCl formed = 2 L (since 1 CH\(_4\) produces 4 HCl)
Step 5: Compute total gaseous volume.
Total gases: \[ \text{CH}_4 (0.5) + HCl (2) = 2.5 \, L \] Cl\(_2\) is fully consumed (2 L used), so no leftover chlorine.
Step 6: Final conclusion.
Thus, total gaseous volume is: \[ \boxed{2.5 \, L} \]
Methane reacts with chlorine to form carbon tetrachloride and hydrogen chloride. Importantly, CCl\(_4\) is a liquid at STP, so it will not contribute to gaseous volume in the final mixture. This is crucial for volume calculation.
Step 2: Write balanced reaction.
\[ CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl \] This shows stoichiometric relationship between reactants and products in gaseous form.
Step 3: Determine amount of CH\(_4\) reacted.
Given 1 L CH\(_4\), 50% reacts: \[ \text{Reacted CH}_4 = 0.5 \, L \] Unreacted CH\(_4\) = 0.5 L remains in gaseous state.
Step 4: Calculate products formed.
From 0.5 L CH\(_4\): - CCl\(_4\) formed = 0.5 L (liquid, not counted in gas volume) - HCl formed = 2 L (since 1 CH\(_4\) produces 4 HCl)
Step 5: Compute total gaseous volume.
Total gases: \[ \text{CH}_4 (0.5) + HCl (2) = 2.5 \, L \] Cl\(_2\) is fully consumed (2 L used), so no leftover chlorine.
Step 6: Final conclusion.
Thus, total gaseous volume is: \[ \boxed{2.5 \, L} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Some basic concepts of chemistry Questions
- The statement related to law of definite proportions is:
- Identify incorrectly matched set from the following
- Arrange the following in the correct order of their bond orders: N$_2$, O$_2$, O$_2^+$, O$_2^-$
- How many grams of Mg is required to completely reduce 100 mL, 0.1 M $\mathrm{NO}_3^-$ solution using the following reaction? $\mathrm{NO}_3^- + \mathrm{Mg} \to \mathrm{Mg}^{2+} + \mathrm{NH}_3$
- Identify the herbicides from the following:
a) DDT b) Aldrin c) Sodium chlorate d) Nicotine e) Sodium arsenite
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions