Ammonium carbonate decomposes upon heating according to the balanced equation:
\(\left(\mathrm{NH}_{4}\right) \mathrm{CO}_{3}(s) \rightarrow 2\ \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)
Calculate the total volume of gas produced at \(22^{\circ} \mathrm{C}\) and 1.02 atm by the complete decomposition of 11.83 g of ammonium carbonate.
Equation Transcription:
Text Transcription:
(NH_4 )CO_3 (s) right arrow 2 NH_3 (g) + CO_2 (g) + H_2 O(g)
22 degrees C
Solution 117P
Given:
(NH4)2CO3(s) → 2NH3(g) + CO2(g) + H2O(g)
Mass of ammonium carbonate = 11.83 g
Pressure = 1.02 atm
Temperature = 22oC
We will have to find out the volume of gas produced.
Explanation:
From the ideal gas equation, it is known that,
PV=nRT
Or, V=(ntotal)RT/P
MW of( NH4)2CO3 (s)=96 gram/mole
moles of ( NH4)2CO3(s)= 11.83 grams of (NH4)2CO3(s)(1 mole of ( NH4)2CO3(s)
------------------------------------------------------------------------ =0.1232
(96 grams)
ntotal of gases produced=moles of NH3 + moles of CO2(g) + moles of H2O (g)
Using the stoichiometry of the balanced eq:
moles of NH3=0.1232 (2)=0.2465
moles of CO2= 0.1232
moles of H2O=0.1232
ntotal of gases produced= 0.1232 +0.1232+ 0.2465=0.4929
273 +Celsius=Kelvin
273 +22 =295 Kelvin
V=(ntotal)RT/P= = 11.71L
Thus the total volume of gas is 11.71 L
