**Answer:**

** 2.12 L **

**Explanation:**

**Data Given:**

Amount of O₂ = 5.3 g

Amount of NH₃ = ?

**Given Reaction:**

** 4NH₃ + 7O₂ ----------> 4NO₂ +6H₂O**

**Solution:**

for details first we look into the reaction

** 4NH₃ + 7O₂ ----------> 4NO₂ + 6H₂O**

** 4 mole 7 mol**

from reaction we come to know that 4 mole of NH₃ react with 7 mole of O₂

**Now we have to convert moles of NH₃ to Liter and mole of O₂ to grams**

** **To convert moles of NH₃ to Liter formula will use

**Volume of NH₃ = no. of moles x molar volume at STP**

**molar volume at STP = 22.4 L/mol**

put values in above equation

**Volume of NH₃ = 4 mol x 22.4 L/mol**

**Volume of NH₃ = 89.6 L **

** Now,**

To convert moles of O₂ to grams

**mass of O₂ = no. of moles x molar mass**

**molar mass of O₂ = 2(16) = 32 g/mol**

Put values in above equation

**mass of O₂ = 7 moles **x** 32 g/mol**

**mass of O₂ = 224 g**

**So,**

We come to know that 89.6 L of NH₃ react with 224 g of O₂ to give NO₂ and water then how many liters of NH₃ will react with 5.3 g O₂

**Now apply unity formula**

** 89.6 L of NH₃ ≅ 224 g of O₂**

** X L of NH₃ ≅ 5.3 g of O₂**

**Do cross multiplication**

** Liter of NH₃ = 89.6 L x 5.3g / 224 g**

** Liter of NH₃ = 2.12 L**

**So**

**option 2 is correct that is 2.12 Liter of NH₃ will react with 5.3 g ** O₂