Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation: CH4(g)+H2O(g) = CO(g) + 3H2(g)
In a particular reaction, 26.5 L of methane gas (measured at a pressure of 736 torr and a temperature of 25 C ) is mixed with 22.8 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 C ). The reaction produces 27.0 L of hydrogen gas measured at STP. What is the percent yield of the reaction?
61.2%
This is about three problems rolled into one. I expect it may be a limiting reagent problem so we address that first.
1. Use PV = nRT to convert CH4 to mols. Remember temperature is in Kelvin and P is in atmospheres.
2. Use PV = nRT to convert H2O to mols. P is in atm and T in Kelvin.
3. Now use the coefficients in the balanced equation and
a. convert mols CH4 to mols H2.
b. convert mols H2O to mols H2.
c. If a and b give the same answer, this is not a limiting reagent problem and you will use that answer throughout the remainder of the calculations. If a and b are not the same, use the smaller of the numbers and that identifies the limiting reagent.
4. Using the appropriate value from 3a,3b, or 3c, convert mols of that compound to mols H2. That number is the theoretical yield of H2 from the reactions under these conditions.
5. Use PV = nRT to convert 27.0 L H2 to mols. This is the actual yield.
6. %yield = [actual yield/theoretical yield]*100 = ??
Check my work carefully. Check my thinking.
Well, it looks like we have a chemical equation party going on here! Let's crunch some numbers and see how this reaction played out.
First, we need to make sure that everything is in the same conditions before the reaction. So, let's convert the pressure and temperature of both reactants to STP (standard temperature and pressure).
For methane gas:
Initial volume = 26.5 L
Initial pressure = 736 torr
Initial temperature = 25°C
To convert the temperature to Kelvin, we add 273.15:
Initial temperature = 298.15 K
Using the ideal gas law, PV = nRT:
n = PV / RT
n = (736 torr × 26.5 L) / (62.36376 torr·L/mol·K × 298.15 K)
n ≈ 1.111 mol
For water vapor:
Initial volume = 22.8 L
Initial pressure = 704 torr
Initial temperature = 125°C
Again, converting the temperature to Kelvin:
Initial temperature = 398.15 K
Using the ideal gas law again:
n = (704 torr × 22.8 L) / (62.36376 torr·L/mol·K × 398.15 K)
n ≈ 0.787 mol
Now, let's see what the balanced equation tells us. We know that 1 mole of methane reacts to produce 3 moles of hydrogen gas, so the theoretical yield of hydrogen gas in this reaction would be:
Theoretical yield = 1.111 mol methane × (3 mol H2 / 1 mol CH4)
Theoretical yield ≈ 3.333 mol H2
But we have experimental data too! We actually measure 27.0 L of hydrogen gas produced at STP.
Since 1 mole of any ideal gas occupies 22.4 L at STP, we can find the actual number of moles of hydrogen gas produced:
Actual yield = 27.0 L / 22.4 L/mol
Actual yield ≈ 1.205 mol H2
Now, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (1.205 mol / 3.333 mol) × 100
Percent yield ≈ 36.16%
So, the percent yield of the reaction is approximately 36.16%. Keep in mind that this is just an estimate, and it's always good to have a sense of humor when dealing with chemistry!
To determine the percent yield of the reaction, we need to compare the amount of hydrogen gas produced in the reaction (actual yield) with the maximum amount of hydrogen gas that could theoretically be produced (theoretical yield).
1. Calculate the number of moles of methane gas (CH4):
Using the Ideal Gas Law, we can convert the given volume of methane gas (26.5 L) to moles.
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin: 25 C + 273.15 = 298.15 K.
Convert the pressure from torr to atm: 736 torr / 760 torr/atm = 0.9684 atm.
Substitute the values into the formula:
(0.9684 atm)(26.5 L) = n(0.0821 atm L/mol K)(298.15 K).
Solve for n to find the number of moles of methane gas.
2. Calculate the number of moles of water vapor (H2O):
Repeat the same steps as above, using the given volume of water vapor (22.8 L) at its respective pressure and temperature.
3. Determine the limiting reactant:
Compare the number of moles of methane gas with the number of moles of water vapor.
The reactant that produces the fewer number of moles of product is the limiting reactant (the one that limits the formation of the product).
4. Calculate the number of moles of hydrogen gas (H2) produced:
Based on the balanced equation, the stoichiometric ratio between methane and hydrogen gas is 1:3.
Determine the number of moles of H2 that can be produced based on the limiting reactant.
5. Calculate the theoretical yield of hydrogen gas:
Convert the number of moles of H2 to liters at STP (Standard Temperature and Pressure) using the ideal gas law:
n = PV / RT, with P = 1 atm and T = 273.15 K.
Convert the number of moles to liters by multiplying by 22.4 L/mol (the molar volume of a gas at STP).
6. Finally, calculate the percent yield:
Divide the actual yield (27.0 L) by the theoretical yield (calculated in step 5), and multiply by 100 to get the percent yield.
Following these steps, you can determine the percent yield of the reaction.