The equation of nitrogen produced in a deployed air bag is: 2NaN3 + 42.7KJ--->2Na + 3N2. What amount of heat is needed to prepare 0.325kg of N2?
42.7 kJ is needed to produce 3*28 = 84 grams. Therefore, to produce 325 g, one will need
42.7 kJ x (325/84) = ??
To determine the amount of heat needed to prepare 0.325 kg of N2 using the given balanced equation, follow these steps:
Step 1: Identify the stoichiometry of the reaction. From the balanced equation, we can see that 2 mol of NaN3 produces 3 mol of N2.
Step 2: Calculate the number of moles of N2 needed. To do this, we convert the mass of N2 to moles using the molar mass of N2, which is approximately 28 g/mol (14 g/mol x 2). First, convert the mass from kg to grams by multiplying by 1000:
0.325 kg * 1000 g/kg = 325 g
Next, calculate the number of moles:
moles of N2 = (mass of N2) / (molar mass of N2)
moles of N2 = 325 g / 28 g/mol ≈ 11.61 mol
Step 3: Determine the heat required. From the balanced equation, it is stated that 2 mol of NaN3 reacts with 42.7 kJ of heat to produce 3 mol of N2. This means that the heat required to produce 1 mol of N2 is:
heat required for 1 mol of N2 = (42.7 kJ) / (2 mol) * (3 mol) ≈ 64.05 kJ
Step 4: Calculate the total heat needed. Multiply the heat required for 1 mol of N2 by the number of moles calculated earlier:
total heat needed = heat required for 1 mol of N2 * moles of N2
total heat needed = 64.05 kJ/mol * 11.61 mol ≈ 745.23 kJ
Therefore, approximately 745.23 kJ of heat is needed to prepare 0.325 kg of N2 using the given equation.