Consider the following equation:
4 NH3 (g) + 5 O2(g) 4 NO (g) + 6 H2O(g)
(a) How many liters of oxygen are required to react with 2.5 L NH3 ? Both gases are at STP.
When gases are involved one may use a short cut in which L are used as if they were mols.
4 NH3 (g) + 5 O2(g) 4 NO (g) + 6 H2O(g)
Therefore, 2.5 L x (5 mols O2/4 mols NH3) = ? L O2
To determine the number of liters of oxygen required to react with 2.5 L of NH3, we need to use the balanced equation and stoichiometry.
The balanced equation is:
4 NH3 (g) + 5 O2(g) -> 4 NO (g) + 6 H2O(g)
From the balanced equation, we can see that 4 moles of NH3 react with 5 moles of O2. However, we are given the volume of NH3 in liters, and we need to convert it to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP is 1 atm)
V = volume (2.5 L)
n = moles (to be determined)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (STP is 273.15 K)
Rearranging the equation, we have:
n = PV / RT = (1 atm)(2.5 L) / (0.0821 L·atm/mol·K)(273.15 K)
Calculating this, we find that the number of moles of NH3 is approximately 0.1172 mol.
Now that we have the number of moles of NH3, we can use the stoichiometry of the balanced equation to determine the number of moles of O2 needed. According to the balanced equation, 4 moles of NH3 react with 5 moles of O2.
Therefore, to find the number of moles of O2, we can set up a ratio:
(0.1172 mol NH3) / (4 mol NH3) = (x mol O2) / (5 mol O2)
Solving for x, we find that x is approximately 0.1465 mol O2.
Finally, we can convert the number of moles of O2 back to liters using the ideal gas law equation. At STP, 1 mole of any gas occupies 22.4 L.
Therefore, the number of liters of O2 required is:
0.1465 mol O2 * 22.4 L/mol = 3.28 L O2
So, approximately 3.28 liters of O2 are required to react with 2.5 L of NH3 at STP.