At high temperatures, solid ammonium nitrate decomposes explosively to form nitrogen gas, water vapour, and oxygen gas according to the following balanced chemical equation:
2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g) 2\text{NH}_4\text{NO}_3(\text{s}) \rightarrow 2\text{N}_2(\text{g}) + 4\text{H}_2\text{O}(\text{g}) + \text{O}_2(\text{g}) 2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g)If 0.80 mol of ammonium nitrate undergoes complete thermal decomposition, how many moles of each gaseous product are formed?
N2=1.60 mol\text{N}_2 = 1.60\text{ mol}N2=1.60 mol, H2O=3.20 mol\text{H}_2\text{O} = 3.20\text{ mol}H2O=3.20 mol, O2=0.80 mol\text{O}_2 = 0.80\text{ mol}O2=0.80 mol
N2=0.80 mol\text{N}_2 = 0.80\text{ mol}N2=0.80 mol, H2O=0.80 mol\text{H}_2\text{O} = 0.80\text{ mol}H2O=0.80 mol, O2=0.80 mol\text{O}_2 = 0.80\text{ mol}O2=0.80 mol
N2=0.80 mol\text{N}_2 = 0.80\text{ mol}N2=0.80 mol, H2O=1.60 mol\text{H}_2\text{O} = 1.60\text{ mol}H2O=1.60 mol, O2=0.40 mol\text{O}_2 = 0.40\text{ mol}O2=0.40 mol
N2=0.40 mol\text{N}_2 = 0.40\text{ mol}N2=0.40 mol, H2O=0.80 mol\text{H}_2\text{O} = 0.80\text{ mol}H2O=0.80 mol, O2=0.20 mol\text{O}_2 = 0.20\text{ mol}O2=0.20 mol