what volume of oxygen gas measured at 25C and 760 torr is required to react with 1.0L of methane measured under the same conditions of temperature and pressure?
CH4 + 2O2 ==> CO2 + 2H2O
You may use L as a shortcut without converting to moles and back to L.
1 mol CH4 will use 2 mols O2.
Would PV=nRT apply to this? if so, how do I find "n"?
Yes, you can use PV = nRT and find n = number of mols. To answer your last question, P = 1 atm (760 torr), V = 1 L, T = 298K (25 C + 273 = 298) and R = 0.08206 L*atm/mol*K.
However, you don't need that.The short cut is that as long as the T and P stay the same (as they do), and you're dealing with gases, you may use L directly and not convert to moles.
CH4 + 2O2 ==> CO2 + 2H2O.
L O2 required = 1L CH4 x (2 moles O2/1 mole CH4) = 1L CH4 x (2/1) = 2L O2 required.
You can prove that to yourself, if you wish, by using PV = nRT as above, solving for n CH4 for the 1.0 L CH4, converting moles CH4 to moles O2, then using PV = nRT to convert moles O2 back to L. The answer is the same by either method but the shortcut is considerably quicker.
To determine the volume of oxygen gas required to react with 1.0L of methane, we need to first write and balance the chemical equation for the reaction between methane (CH4) and oxygen gas (O2).
The balanced equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O
From the balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen gas.
To find the volume of oxygen gas required, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
Given:
Volume of methane (V) = 1.0 L
Temperature (T) = 25°C = 25 + 273.15 = 298.15 K
Pressure (P) = 760 torr
Step 1: Convert temperature to Kelvin:
T = 25°C + 273.15 = 298.15 K
Step 2: Convert pressure to atmospheres:
P = 760 torr/760 = 1 atm
Step 3: Calculate the number of moles of methane:
From the volume and the ideal gas law equation, we can find the number of moles of methane:
n = PV/RT = (1 atm)(1.0 L)/(0.0821 L·atm/(mol·K))(298.15 K)
Step 4: Calculate the number of moles of oxygen gas:
Since 1 mole of methane reacts with 2 moles of oxygen gas, the number of moles of oxygen gas required would be twice the number of moles of methane.
Step 5: Calculate the volume of oxygen gas:
Now that we know the number of moles of oxygen gas, we can calculate its volume using the ideal gas law equation:
V = nRT/P = (2 * number of moles of methane)(0.0821 L·atm/(mol·K))(298.15 K)/(1 atm)
By plugging in the values and performing the calculations, you can find the volume of oxygen gas required to react with 1.0L of methane at 25°C and 760 torr.