# Problems using ∆H(rxn) | StudyDaddy.com

1) Calculate ∆H(rxn) for N2H4(l) + O2(g) = N2(g) 2H2O(l)

​2NH3(g) + 3N2O(g) = 4N2(g) + 3H2O(l) ∆H(rxn) = -1010 kJ/mole

N2O(g) + 3H2(g) = N2H4(l) + H2O(l) ∆H(rxn) = -317 kJ/mole

2NH3 + (1/2) O2(g) = N2H4(l) + H2O(l) ∆H(rxn) = -143 kJ/mole

H2(g) + (1/2) O2(g) = H2O (l) ∆H(rxn) = -286 kJ/mole

2) P4(s) + 6Cl2(g) —-> 4Pcl3(s)  ∆H(rxn) = -1280 kJ/mole

18 grams of P4 react with 42.60g of CL2(g) how much heat will be liberated?

HINT: What is limiting reagent?

3)Calculate ∆H(rxn)for the reaction:

5N2O4(l) +4N2H3CH3(l) = 12H2O(g) + 9N2(g) + 4CO2(g)

Given:

N2H3CH3(l)     ∆H(rxn) = 54 kJ/mole

N2O4(l)            ∆H(rxn) = -20kJ/mole

4)Cup 1=128mL of water at 88ºC

cup2=59.6mL of water at 18.5ºC

What is final temp in ºC when cup #1 is poured into cup 2?

Density of water is 1g/mL

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