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20 ml of nitric oxide combines with 10 m...

20 ml of nitric oxide combines with 10 ml of oxygen at STP to give `NO_(2)`. The final volume will be

A

30 ml

B

20 ml

C

10 ml

D

40 ml

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining the final volume of nitrogen dioxide (NO₂) formed when 20 mL of nitric oxide (NO) combines with 10 mL of oxygen (O₂) at standard temperature and pressure (STP), we can follow these steps: ### Step 1: Write the balanced chemical equation The balanced chemical equation for the reaction between nitric oxide and oxygen is: \[ 2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \] ### Step 2: Determine the initial volumes of reactants We have: - Volume of NO = 20 mL - Volume of O₂ = 10 mL ### Step 3: Calculate the moles of each reactant At STP, the volume of one mole of gas is 22,400 mL. We can calculate the moles of each gas using the formula: \[ \text{Moles} = \frac{\text{Volume}}{\text{Molar Volume}} \] For NO: \[ \text{Moles of NO} = \frac{20 \, \text{mL}}{22,400 \, \text{mL/mol}} = \frac{20}{22,400} \, \text{mol} \] For O₂: \[ \text{Moles of O}_2 = \frac{10 \, \text{mL}}{22,400 \, \text{mL/mol}} = \frac{10}{22,400} \, \text{mol} \] ### Step 4: Determine the limiting reagent From the balanced equation, we see that 2 moles of NO react with 1 mole of O₂. Therefore, the mole ratio is: - 2 moles of NO : 1 mole of O₂ Now, we can find the required moles of O₂ for the available moles of NO: - Required moles of O₂ for 20 mL of NO: \[ \text{Required moles of O}_2 = \frac{10}{22,400} \, \text{mol} \] Now, we can find the moles of NO needed to react with the available O₂: - Available moles of NO for 10 mL of O₂: \[ \text{Required moles of NO} = 2 \times \frac{10}{22,400} = \frac{20}{22,400} \, \text{mol} \] Since we have equal amounts of NO and O₂ in terms of their stoichiometric ratios, both reactants will be completely consumed. ### Step 5: Calculate the moles of NO₂ produced From the balanced equation, 2 moles of NO produce 2 moles of NO₂. Thus, the moles of NO₂ produced will be equal to the moles of NO consumed: \[ \text{Moles of NO}_2 = \text{Moles of NO} = \frac{20}{22,400} \, \text{mol} \] ### Step 6: Calculate the volume of NO₂ produced Using the volume formula: \[ \text{Volume of NO}_2 = \text{Moles of NO}_2 \times \text{Molar Volume} \] \[ \text{Volume of NO}_2 = \frac{20}{22,400} \times 22,400 \, \text{mL} = 20 \, \text{mL} \] ### Final Answer The final volume of nitrogen dioxide (NO₂) produced is **20 mL**. ---
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