Question

Under conditions of `1*atm` pressure, and a temperature of `110` `""^@C`, what volume of gas would decomposition of a `127.0*g` mass of ammonium carbonate evolve?

Answer 1

Well you got the stoichiometric equation.....I make it approx. `160*L`...

Explanation:

`(NH_4)_2CO_3(s)+Delta rarr 2NH_3(g)uarr+CO_2(g)uarr + H_2O(g)uarr`

And thus one mole of ammonium carbonate gives FOUR MOLES of GAS....

`"Moles of ammonium carbonate"=(127.0*g)/(96.09*g*mol^-1)=1.32*mol`

And thus we evolve `4xx1.32*mol` of gas....whose volume we may interrogate by the old Ideal Gas equation.

`V=(nRT)/P=(4xx1.32*cancel(mol)xx0.0821*(L*cancel(atm))/(cancelK*cancel(mol))xx383.0*cancelK)/(1.00*cancel(atm)`

`=??*L`