When heated to 350 degrees C at 0.950 atm, the ammonium nitrate decomposes to pr

Question

When heated to 350 degrees C at 0.950 atm, the ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases; 2NH4NO3(s) delta--->2N2(g)+4H2O(g)+O2(g): a) How many liters of water vapor are produced when 25.8 g of NH4NO3 decomposes? b) How many grams of NH4NO3 are needed to produce 10.0 L of oxygen?

Explanation / Answer

Pressure (P) = 0.0950 atm : temperature (T) = 350C = 623 K : R = 0.0821 L atm/mol K+ 2NH4NO3(s) delta==========>2N2(g)+4H2O(g)+O2(g) a) How many liters of water vapor are produced when 25.8 g of NH4NO3 decomposes? Moles = mass/molar mass Moles 0f NH4NO3 = 25.8g/80.04g/mol=0.322 mol NH4NO3 Now the molar ratio is 2 NH4NO3 : 4 mol H2O 0.322 mol NH4NO3 * ( 4 mol H2O/2 mol 0.322 mol NH4NO3) 0.644 mol H2O according to ideal gas PV=nRT V=nRT/P = (0.644*0.0821*632)/0.0950 = 351.74 L H2O b) How many grams of NH4NO3 are needed to produce 10.0 L of oxygen? according to ideal gas PV=nRT n = PV/RT = (0.0950*10)/(0.0821*632)=0.0183 mol H2O 0.0183 mol H2O * ( 2 mol NH4NO3 / 4 mol H2O)*(80.04g NH4NO3/1 mol NH4NO3) 0.732 g NH4NO3

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