chem help! please!?

Question

i have a whole set of questions to do and i dont understand two of them out of all of them. any help will be great! thanks!<3

47) 2C2H6 + 7O2 ---> 6H2O + 4CO2
A) ----56 grams of C2H6 is combusted. how many liters of H2O will be produced at a pressure of 0.66 atm and 375 K?
B) ----how many liters of O2 will be needed to produce the water?

72) NaNO3 ---> Na2O + N2O5
A)----how much NaNO3 in grams would you need to produce 145.6L of N2O5, at 575 K and 124.6kPa?

Answer 1

The answer depends on whether you can use the ideal gas law, PV=nRT.
A. Solve for the number of moles of H2O gas created after reacting the 56 grams of ethane. Then plug this value, which is "n" , into the ideal gas law with P=.66atm, and T=373K. Find the gas constant with units of L,atm,K, and moles. Then solve for V.

B. Do the same thing again, but use n equal to the number of moles of O2 needed for perfect combustion.

A.Solve the Ideal gas Law for n equal to the number of moles of N2O5. To do this convert your pressure to atm's, then you can use the gas constant from the previous problem. With n you can solve for the number of moles of NaNO3, which can be converted to grams.

Answer 2

Mole wt of c2h6= 2x12+6=28
56gm = 2 moles
2 moles c2h6 yields 6 moles of water
each mole has a volume of 22.5 L at standard conditions of 0 degC and 1 atm. We need to reduce the pressure and increase the temperature--these will both increase the volume.

6 moles x 22.5 L x 1 atm. x 375 degK = 281 L H2O
........................ 0.66atm...273 degK

You understand that I could not add the fraction bar. The above is 1 atm. divided by 0.66 atm. and so on.
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One atmosphere is 0.00987 kPa I am not sure of this.
Mole wt of NaNO3 is 85 see: http://www.chemeurope.com/tools/mm.php3?formel=NaNO3

145.6L x Mole x 273 degK x 124.6 kPa= 38786 gm
............22.5 L...575 degK...0.987 kPa