Question #a7871

1 Answer
Aug 10, 2017

Answer:

Approx......#1*L#........

Explanation:

We gots.......

#NH_4NO_2(s) stackrel(Delta)rarrN_2(g) + 2H_2O(g)#

And thus there is 1:1 molar equivalence between ammonium nitrite, and dinitrogen gas.....

#"Moles of ammonium nitrite"=(2.25*g)/(64.06*g*mol^-1)=0.0351*mol#.

And this we evolve #0.0351*mol# dinitrogen gas......

Now #V=(nRT)/P=(0.0351*mol*0.0821*(L*atm)/(K*mol)xx299.2*K)/((745*mm*Hg)/(760*mm*Hg*atm^-1))#

#~=0.9*L#

The key to answering this problem is to know that #1*atm# pressure will support a column of mercury that is #760*mm# high. And thus we can use the height of a mercury column as a very good VISUAL representation of pressure.