Question

ok heres my question,


nitrogylcerin decomposes:

4c3h4(no5)3 --->12co2(g) +10h20(g)+6n2(g)+02(g)

whats is the max pressure that a 10.0L container will be able to withstand if 5.00 Nitroglycerin decomposes at temp of 1250c

mm of nitroglycerin =227 g/m

i got 2.00atm

b) what are the partial pressures

c02=.828
h20=.690
n2=.414
02=.0690


can u show me the method to do the total pressure. The rest i can do myself

thanks again

Answer 1

See the formula for nitroglycerin here.
Which formula is correct?http://en.wikipedia.org/wiki/Nitroglycerin

Answer 2

the formula is given

4c3h4(no5)3 --->12co2(g) +10h20(g)+6n2(g)+02(g)

Answer 3

Your equation isn't balanced. 16 H on the left (and you should try to capitalize those letters that need it) and I see 20 on the right. 60 O on the left and 36 on the right.

Answer 4

it was given so we don't have to. I think the teacher is just trying to show us how to use the mole ratios to get partial pressures

Answer 5

here this might help

4 C3H5(NO3)3(l) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)

I must have copied it wrong into the browser.



sorry...