N2H4 reacts with oxygen to yield nitrogen gas and water . The reaction of 3.80g n2h4 evolves 73.7kj of heat. what's the enthalpy change per mole of hydrazine combusted.
i forgot to add the equation: n2H4 + 02 -> N2 + h2o
You know the enthalpy change per 3.80 grams. You want to know the change per 32 (1 mole) grams.
73.7 kJ x (32/3.80) = ??
To find the enthalpy change per mole of hydrazine combusted, we need to first convert the given mass of N2H4 to moles.
1. Calculate the molar mass of N2H4:
N = 14.01 g/mol (atomic mass of nitrogen)
H = 1.01 g/mol (atomic mass of hydrogen)
Molar mass of N2H4 = 2(N) + 4(H)
= 2(14.01) + 4(1.01)
= 28.02 + 4.04
= 32.06 g/mol
2. Convert the given mass of N2H4 to moles:
Moles = Mass / Molar mass
= 3.80 g / 32.06 g/mol
= 0.1185 mol
3. Calculate the enthalpy change per mole of N2H4:
Enthalpy change per mole = Heat evolved / Moles
= 73.7 kJ / 0.1185 mol
= 621.77 kJ/mol
Therefore, the enthalpy change per mole of hydrazine combusted is 621.77 kJ/mol.