Question

How much energy is given off by the following reaction, if 162.5 g of oxygen reacts with

216.7 g of ammonia (NH₃)?
4NH₃ + 5O₂ ⇌ 4NO + 6H₂O
AH = -1225.6 kJ
A) +1245 kJ
B) +39830 kJ
C) -39830 kj
D) -1245 k)

Answer 1

The correct answer is option C) -39830 kj.

Step-by-step explanation:

To determine the energy given off by the reaction, we need to calculate the number of moles of reactants and use the stoichiometric coefficients to find the moles of products.

Given:

• Mass of oxygen (O₂) = 162.5 g
• Mass of ammonia (NH₃) = 216.7 g
• Equation: 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O
• Standard enthalpy change (ΔH°) = -1225.6 kJ

First, we calculate the moles of oxygen and ammonia using their molar masses:

• Molar mass of oxygen (O₂) = 32 g/mol
• Moles of oxygen = 162.5 g / 32 g/mol = 5.08 mol
• Molar mass of ammonia (NH₃) = 17 g/mol
• Moles of ammonia = 216.7 g / 17 g/mol = 12.75 mol

According to the balanced equation, 5 moles of oxygen react with 4 moles of ammonia to produce 4 moles of nitric oxide (NO) and 6 moles of water (H₂O).

Using this ratio, we can calculate the moles of products:

• Moles of nitric oxide (NO) = (5.08 mol O₂) × (4 mol NO / 5 mol O₂) = 4.064 mol
• Moles of water (H₂O) = (12.75 mol NH₃) × (6 mol H₂O / 4 mol NH₃) = 19.125 mol

Finally, we can calculate the energy given off by the reaction using the stoichiometric coefficients and the standard enthalpy change:

• Energy given off = ΔH° × (moles of nitric oxide + moles of water) = -1225.6 kJ × (4.064 mol + 19.125 mol) = -39829.68 kJ.

Therefore, the amount of energy given off by the reaction is approximately -39830 kJ.