How many liters of gas would be formed at 455 degrees C and 1.00 atm pressure by

Question

How many liters of gas would be formed at 455 degrees C and 1.00 atm pressure by explosion of 540g of NH4NO3? 2NH4NO3 (s) ----------> 2N(g) + 4H2O (g) + O2 (g)

Explanation / Answer

This is an ideal gas law problem. That law states that: PV = nRT P - pressure (atm) V - volume (L) n - moles (mol) R - gas constant, 0.08206 (L*atm)/(mol*K) T - temperature (K) So the reaction you're looking at is ammonium nitrate, which is converted to nitrogen, water, and oxygen gas during the reaction. Let's simplify the way that equation looks (keeping in mind that the products are gases): 2NH4NO3 ---> 2N2 + 4H2O + O2 Since reactions happen at the molecular level, let's convert the amount of reactant given in (g) to moles. Then we can use a mole ratio to determine exactly how much gas was produced. First you will need to calculate the molar mass of the ammonium nitrate (I assume you know how to do that). 550g NH4NO3 (1 mol / _____ g NH4NO3) = ? moles of NH4NO3 Obviously, this reaction is "explosive" because two moles of reactant produce 7 moles of gas, and that's a significant expansion in volume. We can say that mathematically as a mole ratio: 2 mol NH4NO3 / ( 2 mol N2 + 4 mol H2O + 1 mol O2) 2 mol NH4NO3 / 7 mol gaseous products Now we can use this mole ratio to calculate how many moles of gaseous products were created. Then we can use the ideal gas law to determine how much volume (in L) that amount of gas would take up under the stated conditions. ? moles of NH4NO3 ( 7 mol gaseous products / 2 mol NH4NO3) = ?? moles of gaseous products That number of moles is "n" in the ideal gas law. And the problem gives us the other variables: T = 495C = convert this to Kelvin (K = C + 273.15) P = 1.00atm So we just have to plug in the numbers and solve for V: PV = nRT V = nRT/P V = [(?? moles of gaseous products)(0.08206 L*atm/mol*K)(T in Kelvin)] / (1.00atm)

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