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10 ml N(2) is reacted with 20 ml H(2) to...

`10 ml N_(2)` is reacted with 20 ml `H_(2)` to form `NH_(3)`. The correct statements is /are

A

13.3 ml `NH_(3)` is formed

B

20 ml `NH_(3)` is formed

C

3.4 ml `N_(2)` is left after the completion of the reaction

D

16.7 ml of mixture is left after the completion of the reaction.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia (NH₃). The balanced chemical equation for the reaction is: \[ \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \] ### Step 1: Identify the volumes of reactants We are given: - Volume of N₂ = 10 ml - Volume of H₂ = 20 ml ### Step 2: Determine the limiting reagent From the balanced equation, we know that 1 ml of N₂ reacts with 3 ml of H₂. Therefore, for 10 ml of N₂, the amount of H₂ required is: \[ 10 \, \text{ml N}_2 \times 3 \, \text{ml H}_2/\text{ml N}_2 = 30 \, \text{ml H}_2 \] However, we only have 20 ml of H₂ available. This means that H₂ is the limiting reagent. ### Step 3: Calculate the amount of N₂ consumed Since H₂ is limiting, we need to find out how much N₂ will react with the available 20 ml of H₂. According to the stoichiometry of the reaction: \[ 3 \, \text{ml H}_2 \text{ reacts with } 1 \, \text{ml N}_2 \] So, for 20 ml of H₂: \[ 20 \, \text{ml H}_2 \times \frac{1 \, \text{ml N}_2}{3 \, \text{ml H}_2} = \frac{20}{3} \, \text{ml N}_2 \approx 6.67 \, \text{ml N}_2 \] ### Step 4: Calculate the remaining N₂ Initially, we had 10 ml of N₂. After the reaction, the amount of N₂ left is: \[ 10 \, \text{ml N}_2 - 6.67 \, \text{ml N}_2 = 3.33 \, \text{ml N}_2 \approx 3.4 \, \text{ml N}_2 \] ### Step 5: Calculate the amount of NH₃ produced From the balanced equation, we can see that 3 ml of H₂ produces 2 ml of NH₃. Therefore, for 20 ml of H₂: \[ 20 \, \text{ml H}_2 \times \frac{2 \, \text{ml NH}_3}{3 \, \text{ml H}_2} = \frac{40}{3} \, \text{ml NH}_3 \approx 13.33 \, \text{ml NH}_3 \] ### Step 6: Calculate the total volume of the gas mixture after the reaction After the reaction, we have: - NH₃ produced = 13.33 ml - Remaining N₂ = 3.33 ml Thus, the total volume of the gas mixture after the reaction is: \[ 13.33 \, \text{ml NH}_3 + 3.33 \, \text{ml N}_2 = 16.66 \, \text{ml} \approx 16.7 \, \text{ml} \] ### Conclusion Now we can evaluate the statements: - **Statement A:** 13.3 ml NH₃ is formed. **(Correct)** - **Statement B:** 20 ml NH₃ is formed. **(Incorrect)** - **Statement C:** 3.4 ml N₂ is left after the completion of the reaction. **(Correct)** - **Statement D:** 16.7 ml of mixture is left after the completion of the reaction. **(Correct)** ### Final Answer The correct statements are A, C, and D.
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