Question

Question #a7871

Answer 1

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.