Ammonium nitrite undergoes decomposition to produce only gases:
NH4NO2 (s) -> N2(g) + H2O (g)
How many liters of gas will be produced by the decomposition of 32.0g of NH4NO2 at 525C and 1.5 atm?
My approach is this:
finds the mole (n) of one of the gases (let say N2)
32g NH4NO2 x (1 mole / molar mass of NH4NO2) x (1 mole N2 / 1 mole NH4NO2)
This stioche will gives us the mole of N2 presents
Now I will use the gase law forumla PV = nRT to find V (N2)
So N2 = x amount in liters
but since there are more than one gases, so I have to do similar setup for H2O (g)
My question is, am I correct about the approach,
that the answer (liter) is equal to N2 (v) + H2O (v).
In general,
2 NaN3(s) --> 2Na(l) + 3N2(g) only one gas so I only find one
but in that question, I have two gases, H2O and N2.
So the total volume is = N2+H2O. am I correct? or am I wrong?
Thanks