A mixture of ammonia and oxygen is prepared by combining 0.330 L of NH3 (measured at 0.710 atm and 22°C) with 0.200 L of O2 (measured at 0.740 atm and 51°C).
How many milliliters of N2 (measured at 0.740 atm and 100. °C) could be formed if the following reaction occurs?
4NH3(g) + 3O2(g) -----> 2N2(g) + 6H2O(g)
figure the moles of NH3 and O2 at the temp, pressures given.
THen the ratio in the balanced equation is for every mole of NH3, you have to have .75 moles of O2. Do you have enough or too much O2? If too much, then the reacdtion is limited by NH3, and you get .5 moles of N2 for each NH3 mole.
If you have too little O2, then yo will get for each O2 mole, 2/3 mole of N2
then convert the moles of N2 to volume using the ideal gas law.
To determine the number of milliliters of N2 that can be formed, we need to use the given quantities of the reactants (NH3 and O2) to determine the limiting reactant.
1. Convert the given volumes of NH3 and O2 to moles at standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atm.
For NH3:
- Use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L•atm/mol•K), and T is the temperature in Kelvin.
- Convert the temperature from Celsius to Kelvin: 22°C + 273 = 295 K.
- Solve for the number of moles of NH3: n = (P * V) / (R * T) = (0.710 atm * 0.330 L) / (0.0821 L•atm/mol•K * 295 K) = 0.00875 mol.
For O2:
- Convert the temperature from Celsius to Kelvin: 51°C + 273 = 324 K.
- Solve for the number of moles of O2: n = (P * V) / (R * T) = (0.740 atm * 0.200 L) / (0.0821 L•atm/mol•K * 324 K) = 0.00528 mol.
2. Use the stoichiometry of the balanced equation to determine the limiting reactant.
Based on the balanced equation:
4 moles of NH3 react with 3 moles of O2 to produce 2 moles of N2.
Let's determine the number of moles of N2 that can be formed from the given amounts of NH3 and O2:
- Calculate the moles of N2 that can be formed from NH3: 0.00875 mol of NH3 * (2 mol of N2 / 4 mol of NH3) = 0.00437 mol of N2.
- Calculate the moles of N2 that can be formed from O2: 0.00528 mol of O2 * (2 mol of N2 / 3 mol of O2) = 0.00352 mol of N2.
Since the moles of N2 that can be formed from O2 (0.00352 mol) are less than the moles of N2 that can be formed from NH3 (0.00437 mol), O2 is the limiting reactant.
3. Convert the moles of N2 to volume using the ideal gas law.
Assuming the temperature and pressure are consistent with the conditions given for N2, which are 100°C and 0.740 atm:
- Convert the temperature from Celsius to Kelvin: 100°C + 273 = 373 K.
- Solve for the volume of N2: V = (n * R * T) / P = (0.00352 mol * 0.0821 L•atm/mol•K * 373 K) / 0.740 atm = 1.06 L = 1060 mL.
Therefore, approximately 1060 milliliters of N2 could be formed.