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7. How many liters of NH3, at STP, will react with 5.3 g 02 to form NO2 and water? (1 point)

4NH, (g)+70,(8) -4N0, +6H,0 (g)
0.00423 L
02.12 L
03.03 L
6.49 L

User Bryanjez
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1 Answer

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Answer:

2.12 L

Step-by-step 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

User Dahlgren
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5.6k points