How many litres of oxygen are needed to react completely with 25.0L of H2S at STP according to the following reaction: 2H2S + 3o2 --> 2so2 + 2H2O
At STP temperature is 273 K pressure is 101.3 Kpa R is 8.314 so PV=nRT
101.3×25= n× 8.314× 273
n= 1.11 so using the molar ratio we find the moles of o2= 1.70 now how do I find the volume of O2 with only the moles and no concentration?!
YOu are doing it the very hard way. All the reactants are gases, same temp, pressure, so you can use the EQual Volumes law (John Dalton)
so for each 2 volumes of H2S, you need three volumes of O2
VolumesO2=3/2 * 25 liters
I agree with Bob Pursley that you are/were doing it the hard way; however, the answer to your question is this. After you know mols O2 = 1.70, then you know that at STP the volume of any gas is 22.4 L for one mole.
So 1.70 mols x (22.4 L/mol) = ? L.
To find the volume of oxygen (O2) needed, you need to use the ideal gas law equation, PV = nRT, which relates the pressure, volume, number of moles, temperature, and the ideal gas constant.
Here's how you can calculate the volume of oxygen:
1. Start with the number of moles of oxygen you calculated, which is 1.70 moles (as you mentioned).
2. Use the molar ratio from the balanced equation to determine the moles of oxygen. The balanced equation shows that 2 moles of H2S react with 3 moles of O2.
Since you have 1.70 moles of O2, you can set up the following ratio:
1.70 moles O2 / 3 moles O2 = X moles H2S / 2 moles H2S
Solve the above ratio for X (moles H2S) by cross-multiplying:
X = (1.70 moles O2 * 2 moles H2S) / 3 moles O2
X = 1.13 moles H2S
3. Now, consider the ideal gas law equation PV = nRT. Rearrange it to solve for volume (V):
V = (nRT) / P
Plug in the values:
V = (1.13 moles * 8.314 * 273 K) / 101.3 KPa
V = 25.6 liters
Therefore, the volume of oxygen needed to react completely with 25.0 liters of H2S at STP is approximately 25.6 liters.