Chemistry | Enthalpy Change Calculation for N2H4 Combustion

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SUMMARY

The discussion focuses on calculating the enthalpy change for the combustion of hydrazine (N2H4) when it reacts with oxygen to produce nitrogen gas and water. The reaction is represented as N2H4 (l) + O2 (g) → N2 (g) + 2H2O (l). The combustion of 6.50 grams of hydrazine releases 126.2 kJ of heat, leading to the need for calculating the enthalpy change per mole of hydrazine combusted. The key step involves converting the mass of hydrazine into moles to facilitate this calculation.

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Homework Statement

hydrzine, N2H4 ia a liquid rocket feul. It reacts with oxygen to yeild nitrogen gas and water.

N2H4 (l)+ O2 (g)--> N2 (g) + 2H2O (l)

the reaction of 6.50g N2H4 evolves 126.2 kJ of heat. Calculate the enthalpy change per mole of hydrazine combusted

I am having trouble getting started on this problem. any help with how to solve this problem will be appreciated. I am having trouble with this section :/

thanks,

Joe
 
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This is just about conversion between mass and number of moles.

6.50 g of hydrazine - how many moles?
 

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