NH4NO3 --> N2O + 2H2O
you start with 1000g of ammonium nitrate, what is the theoretical yield of laughing gas? How many grams of water will be produced.
*I know you COULD have 450.234g of H20, but I'm not sure how to do the theoretical yield.
The equation is balanced. Compute #mols in 1000 g NH4NO3, convert to mols N2O (the same since 1 mol NH4NO3 produces 1 mol N2O) and from that grams. Do the same thing for water except two mols H2O are produced for each mol of NH4NO3. Those two numbers will be the theoretical yield for N2O and H2O respectively. You need to clarify the 450.234 g H2O you have posted. I don't know what that is.
ignore that "2"
To calculate the theoretical yield of laughing gas (N2O) and the amount of water (H2O) produced from the reaction of 1000g of ammonium nitrate (NH4NO3), we first need to determine the limiting reactant.
1. Balanced chemical equation:
NH4NO3 -> N2O + 2H2O
2. Determine the molar masses of the compounds involved:
NH4NO3: 14.01 + 4(1.01) + 3(16.00) = 80.04 g/mol
N2O: 2(14.01) + 16.00 = 44.02 g/mol
H2O: 2(1.01) + 16.00 = 18.02 g/mol
3. Calculate the number of moles of ammonium nitrate (NH4NO3) present:
moles of NH4NO3 = mass / molar mass
moles of NH4NO3 = 1000g / 80.04 g/mol ≈ 12.498 mol
4. Use stoichiometry to determine the molar ratio between NH4NO3 and N2O:
From the balanced equation, the ratio is 1:1. Therefore, the number of moles of N2O produced will be the same as the number of moles of NH4NO3 used.
5. Calculate the mass of N2O produced:
mass of N2O = moles of N2O x molar mass
mass of N2O = 12.498 mol x 44.02 g/mol ≈ 549.949 g
So, the theoretical yield of laughing gas (N2O) is approximately 549.949 grams.
6. Calculate the mass of water (H2O) produced:
From the balanced equation, for every 1 mole of NH4NO3, 2 moles of water are produced.
Therefore, the moles of water produced will be twice the moles of ammonium nitrate used.
moles of H2O = 2 x moles of NH4NO3
moles of H2O = 2 x 12.498 mol ≈ 24.996 mol
mass of H2O = moles of H2O x molar mass
mass of H2O = 24.996 mol x 18.02 g/mol ≈ 450.432 g
So, the amount of water (H2O) produced is approximately 450.432 grams.