Liquid hydrogen peroxide, an oxidizing agent in many rocket fuel mixtures, releases oxygen gas on decomposition.
2 H2O2(l) 2 H2O(l) + O2(g) Hrxn = -196.1 kJ
How much heat is released when 529 kg H2O2 decomposes?
( ......... ) KJ
*note ..watch sig figs .
my work:
• 2 H2O2(l) => 2 H2O(l) + O2(g) ΔHrxn = -196.1 kJ
Moles of H2O2 = mass/molar mass
= 529,000/34 = 15558.82 moles of H2O2 release 196.1 kJ of heat
Total heat released = 15558.82/2 x196.1 = 1525542.301 = 1.52554 × 10^6
= 1.53 x 10^6 kJ (3 significant figures)
am I correct?.....thank you.
Your calculations are almost correct, but there is a slight mistake in the final step of your calculation.
To calculate the heat released when 529 kg of H2O2 decomposes, you need to use the balanced equation and the given enthalpy change:
2 H2O2(l) -> 2 H2O(l) + O2(g) ΔHrxn = -196.1 kJ
First, calculate the moles of H2O2:
Moles of H2O2 = mass / molar mass
Molar mass of H2O2 = 1 * 2 (H) + 2 * 16 (O) = 34 g/mol
Mass of H2O2 = 529 kg = 529,000 g
Moles of H2O2 = 529,000 g / 34 g/mol = 15,558.82 moles
Now, calculate the heat released:
Total heat released = (moles of H2O2 / 2) * ΔHrxn
= (15,558.82 moles / 2) * -196.1 kJ/mol
= 7,779.41 moles * -196.1 kJ/mol
= -1,522,297.71 kJ (rounded to five significant figures)
So, the correct answer is -1,522,298 kJ (to three significant figures) or -1.52 x 10^6 kJ.
Your final answer is almost correct, but you rounded off too early. The correct answer should have five significant figures, so it would be -1,522,298 kJ. Remember to round after the full calculation and not during the intermediate steps.