Still have no clue how to do thise?

Question

Only the 2nd part... Sorry for bugging

Even with the fine help you guys have given me, someone please explain!

N2H4 (g) + O2 (g) --> N2 (g) + 2H20 (I) Delta H=-628 KJ

How many grams of Hydrazine (N2H4) would be required to produce 963 KJ of heat?

If the reaction was at standard temperature and pressure, how many liters of N2 gas would evolve if the amount of N2H4 found in part (a) was burned?

Answer 1

Hope you understand I am not goin to solve this for you, It would be unethical, but here is what you need to do:

Think this; ONE mol of hydrazine generates 628 KJ So, make yourself this question: How many (x) moles of hydrazine will generate 963 KJ

SEE:

http://en.wikipedia.org/wiki/Rule_of_three_%28mathematics%29

come on, you can do it!

Then , when you get the moles of hydrazine, it will be easy to transform it into grams, multiplying it by its molecular mass as follows:

moles=mass/molar mass

then

mass = moles x molar mass

If you haven't calculated the molar mass of hydrazine yet proceed as follows: sum the atomic mass of all its atoms, two nitrogens and four hydrogens. Or find it at:

http://en.wikipedia.org/wiki/Hydrazine


For the last question:

Think of this: Observe in the reaction how many molecules of N2 are generated by one molecule of hydrazine. ONE, right? So mols of nitrogen = mols of hydrazine

then take the perfect gas law,
PV=nRT

rearrange:

V=nRT/P

sustitute with the appropiate units and there you have it!
remember T must be in Kelvin (Not celsius, neither Farenheith)

Look here which are the standard temperature and pressure:

http://en.wikipedia.org/wiki/Standard_Temperature

Which is the right R value for you? Find it here;

http://en.wikipedia.org/wiki/Gas_constant


That's all I can dare to do for you. Hope you learn something on the way. Good luck

Answer 2

If one mole of N2H4 produces -628 KJ then it takes 963/628 moles to produce 963 KJ.

1 mole of N weighs 14, 1 mole of H weighs 1
1 moles of N2H4 weighs 2*14+4*1 = 30 grams

963/628 = 1.53343949 moles

Ans first question is 963/628 * 32 = 49.070 grams

2nd question. Use Idea Gas Equation
PV=NRT: R = .082056 liter atm/(K Mol)
and STP => P = 1 atm, T=273.15 K,

Volume = 1.533 mol * .082056 * 273.15 K/1 ATM
= 34.37 liters

Answer 3

Do it as moles. Molar enthalpy of reaction, molar volume of gas. Moles to grams and back are trivial.