Question

If there are 2.36 g of NO present, how much energy (kJ) will be produced?

2 NO (g) → 2 N2 (g) + O2 (g) ΔH = -180.0 kJ

If 3.50 kJ was required for the following reaction, how many grams of KClO₃ was produced?

2 KCl (s) + 3 O₂ (g) → 2 KClO₃ (s)   ΔH = 84.9 kJ

If there are 0.709 mol of O₂ present, how much energy (kJ) will be consumed when the reaction is reversed?

H₂O₂ (l) → H₂O (l) + O₂ (g) ΔH = -196.4 kJ

Answer 1

2 NO (g)----> 2 N2 (g) + O2 (g) dH = -180.0 kJ

it is given that 2 mole NO to produce product with the
release of 180 KJ energy.negative dH value indicate reaction
is exothermic

So 1 mole NO release 180/2 Kj =90 KJ of energy
2.36 g of NO = 2.36/molar mass of NO moles of energy
= 2.36 g/30.01 g/mol = 0.0786 moles
SO there is 0.0786 moles of NO

0.0786 moles of NO release energy = 90 kj*0.0786
= 7.078 KJ
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2 KCl (s) + 3 O2 (g)----> 2 KClO3 (s) dH = 84.9 kJ

dH is positive means reaction is endothermic ; that means it absorbs energy
From balanced equation we can make out that 2 moles KClO3 produced when
84.9 KJ energy is absorbed

1 KJ energy released when 2/84.9 mole KClO3 produced
3.50 KJ is released when [2/84.9]* 3.50 mole KClO3
produced = 0.0824 moles

moles of KClO3 produced = 0.0824 moles

convert this moles to mass using molar mass of KCLO3

mass of KClO3 produced = # moles of KClO3*molar mass
= 0.0824 mole*122.55 g/mol = 10.10 g
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H2O2 (l)----> H2O (l) + O2 (g) dH = -196.4 kJ

the given equation says 1 mole O2 produced
along with the release of 196.4 kJ of heat
So when 0.709 mole O2 produced ;
energy released = 0.709 * 196.4 KJ = 139.2476 KJ
dH = - 139.2476 KJ
This much energy will be released
When this reaction is reverse same amount of heat will be
absorbed.
for reverse reaction keep in mind that ,since it is
endotermic,sign will be positive
SO dH = +139.2476 KJ
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Thank you .hope it is helpful