When heated to 350. ∘C at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases:
2NH4NO3(s)→Δ2N2(g)+4H2O(g)+O2(g)
Using the ideal gas law equation, calculate the volume, in liters, of water vapor produced when 21.8 g of NH4NO3 decomposes.
1. mols NH4NO3 = grams/molar mass = ?
2. Using the coefficients in the balanced equation, convert mols NH4NO3 to mols H2O.
3. Now use PV = nRT, substitute mols from above for n, use the conditions of P and T in the problem, solve for V in liters. Remember to change 350 C to kelvin.
Post your work if you get stuck.
im very stuck on finding r
Does V=188.4 L?
Just kidding. I mean 51.34L?
To calculate the volume of water vapor produced when 21.8 g of NH4NO3 decomposes, we need to follow these steps:
1. Convert the mass of NH4NO3 to moles.
2. Use stoichiometry to determine the moles of water vapor produced.
3. Apply the ideal gas law equation to calculate the volume of water vapor in liters.
Let's go through each step more detailed:
Step 1: Convert the mass of NH4NO3 to moles.
To convert grams to moles, we need to divide the given mass by the molar mass of NH4NO3.
The molar mass of NH4NO3 is:
(N = 14.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol)
(2 × N + 4 × H + 3 × O) = (2 × 14.01 g/mol + 4 × 1.01 g/mol + 3 × 16.00 g/mol) = 80.05 g/mol
We can now calculate the moles of NH4NO3:
Moles of NH4NO3 = Mass of NH4NO3 / Molar mass of NH4NO3
Moles of NH4NO3 = 21.8 g / 80.05 g/mol
Step 2: Use stoichiometry to determine the moles of water vapor produced.
From the balanced equation, we can see that 2 moles of NH4NO3 produce 4 moles of H2O.
Since we have the moles of NH4NO3, we can use stoichiometry to calculate the moles of H2O:
Moles of H2O = (Moles of NH4NO3) × (4 moles of H2O / 2 moles of NH4NO3)
Step 3: Apply the ideal gas law equation to calculate the volume of water vapor in liters.
The ideal gas law equation is: PV = nRT
We have the following information:
Pressure (P) = 0.950 atm (given)
Temperature (T) = 350 °C = 350 + 273.15 K = 623.15 K
Gas constant (R) = 0.0821 L·atm/(mol·K) (constant)
Moles of H2O (n) = (Moles of H2O) from Step 2
Rearranging the ideal gas law equation, we get:
V = nRT / P
Now we can calculate the volume of water vapor:
Volume of H2O = (Moles of H2O) × (R) × (T) / P
Using the given values, plug them into the formula and solve for the volume of water vapor in liters.