Question

10 Na (5) + 2 KNO, (s) + K,0 (s) + 5 Na, o (s) + N, (9) AH = -1457.20 kJ • Calculate the overall change in enthalpy when 18.00 grams of Na (5) reacts with excess KNO, (s). [Be sure to include the correct sign.] Enthalpy change: • How many grams of Na, o (s) are produced when the reaction above releases 271.5 kJ of heat energy. grams of Na, o (s) Check

Answer 1

Answer:

Step 1: Calculate the moles from mass

We know, moles = mass / molar mass

(1) moles of Na(s) = 18 g / 22.99 g/mol = 0.78295 mol

Step 2: Calculation

(1) Write the balance equation

10 Na(s) + 2 KNO₃(s)----> K₂O(s) + 5 Na₂O(s) + N₂(g) ; ΔH=−1457.20 kJ

For 10 moles of Na(s) enthaply change = -1457.20 kJ

so, for 0.78295 mol of Na(s) enthapy change will be = ( -1457.20 kJ / 10 mol ) × 0.78295 mol = -114.1 kJ

Step 3: (part 2)

10 Na(s) + 2 KNO₃(s)----> K₂O(s) + 5 Na₂O(s) + N₂(g) ; ΔH=−1457.20 kJ

From the reaction we can see that

The releases of enthalpy of 1457.20 kJ when 5 moles of Na₂O(s) produced

so, for the releases of enthalpy of 271.5 kJ   Na₂O(s) moles produced   = ( 5 mol / 1457.20 kJ  ) × 271.5 kJ = 0.9316 mol

Now convert mole into mass of Na₂O(s)

We know, mass = moles × molar mass of Na₂O(s) = 0.9316 mol × 61.9789 g/mol = 57.7 g

Answer:

1. -114.1 kJ

2. 57.7 g