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10 ml of a mixture of carbon monoxide, m...

10 ml of a mixture of carbon monoxide, marsh gas and hydrogen exploded with excess of oxygen gave a contraction of 6.5 CC. There was further contraction of 7 CC when the residual gas was treated with caustic potash. The volume of marsh gas present in original mixture is

A

5CC

B

2CC

C

3CC

D

4CC

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will define the variables and set up equations based on the information provided in the question. ### Step 1: Define Variables Let: - \( x \) = volume of carbon monoxide (CO) in ml - \( y \) = volume of marsh gas (methane, CH₄) in ml - \( z \) = volume of hydrogen (H₂) in ml From the problem, we know that the total volume of the mixture is 10 ml. Therefore, we can write our first equation: \[ x + y + z = 10 \quad \text{(Equation 1)} \] ### Step 2: Analyze the Reaction with Oxygen When the mixture is exploded with excess oxygen, the following reactions occur: 1. Carbon monoxide reacts with oxygen to form carbon dioxide: \[ \text{CO} + \text{O}_2 \rightarrow \text{CO}_2 \] For \( x \) ml of CO, it will react with \( \frac{x}{2} \) ml of O₂ and produce \( x \) ml of CO₂. 2. Marsh gas (methane) reacts with oxygen to form carbon dioxide and water: \[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \] For \( y \) ml of CH₄, it will react with \( 2y \) ml of O₂ and produce \( y \) ml of CO₂. The total contraction in volume due to the consumption of oxygen is given as 6.5 ml. The contraction is due to the volume of oxygen consumed, which can be expressed as: \[ \frac{x}{2} + 2y = 6.5 \quad \text{(Equation 2)} \] ### Step 3: Analyze the Contraction with Caustic Potash After the explosion, the residual gas is treated with caustic potash, which absorbs carbon dioxide. The further contraction is given as 7 ml. This means the total volume of carbon dioxide produced from the reactions is equal to the contraction: \[ x + y = 7 \quad \text{(Equation 3)} \] ### Step 4: Solve the Equations Now we have a system of three equations: 1. \( x + y + z = 10 \) (Equation 1) 2. \( \frac{x}{2} + 2y = 6.5 \) (Equation 2) 3. \( x + y = 7 \) (Equation 3) From Equation 3, we can express \( x \) in terms of \( y \): \[ x = 7 - y \] Now substitute \( x \) in Equation 2: \[ \frac{7 - y}{2} + 2y = 6.5 \] Multiply through by 2 to eliminate the fraction: \[ 7 - y + 4y = 13 \] Combine like terms: \[ 7 + 3y = 13 \] Subtract 7 from both sides: \[ 3y = 6 \] Divide by 3: \[ y = 2 \] ### Step 5: Find the Volume of Marsh Gas Now that we have \( y \), we can find \( x \): \[ x = 7 - y = 7 - 2 = 5 \] ### Conclusion The volume of marsh gas (methane) present in the original mixture is: \[ \boxed{2 \text{ ml}} \]
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