Hydrogen gas is produced industrially by reacting methane gas, CH4 (g), with water vapor to from carbon monoxide and hydrogen gas. In a particular reaction, 25.5L of methane (measured at a pressure of 732 torr and a temperature of 25degree Celcius) mixes with 22.8 L of water vapor(measured at a pressure of 702 torr and a temperature of 125 degree Celcius). What is the theoretical yield hydrogen gas at STP? Give your answer in units of L to three significant figures.
write the balanced equation:
CH4+H2O >> 3H2 + CO
get all the reactants moles:
CH4: n=PV/RT=732*25.5/R(298)=62.6/R
H2O: n=22.8*702/R(398)=40.2/R
so you have an excess of CH4.
So according to the balanced equation, you will get moles of H2 of 3*40.2/R =121/R moles H2
of volume determined by
V=nRT/P=121/R * R*273/760=43.7L
check my work.
To find the theoretical yield of hydrogen gas at STP (Standard Temperature and Pressure), we need to determine the number of moles of methane and water vapor and then use the balanced equation to calculate the number of moles of hydrogen gas produced. Since the pressure and temperature of both gases are given, we need to use the ideal gas law equation to find the number of moles.
First, let's find the number of moles of methane:
Using the ideal gas law equation, 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.
For methane:
P = 732 torr = 732/760 atm (since 1 atm = 760 torr)
V = 25.5 L
T = 25 degrees Celsius = 25 + 273.15 K (converting to Kelvin)
Plugging the values into the ideal gas law equation:
732/760 * 25.5 = n * 0.0821 * (25 + 273.15)
Simplifying the equation and solving for n(moles of methane):
n = (732/760 * 25.5) / (0.0821 * 298.15)
Now, calculate the number of moles of water vapor:
Using the same procedure as above, but with the given data for water vapor:
P = 702 torr = 702/760 atm
V = 22.8 L
T = 125 degrees Celsius = 125 + 273.15 K
Plugging the values into the ideal gas law equation:
702/760 * 22.8 = n * 0.0821 * (125 + 273.15)
Simplifying the equation and solving for n(moles of water vapor):
n = (702/760 * 22.8) / (0.0821 * 398.15)
Now, we need to use the balanced equation to find the molar ratio between methane and hydrogen gas.
The balanced equation for the reaction is:
CH4 (g) + H2O (g) → CO (g) + 3H2 (g)
According to the balanced equation, for every mole of methane, 3 moles of hydrogen gas are produced.
Therefore, the number of moles of hydrogen gas produced is:
3 * (moles of methane)
Now, we need to convert the moles of hydrogen gas to liters at STP (Standard Temperature and Pressure), where STP conditions are 0 degrees Celsius and 1 atm of pressure. The molar volume at STP is 22.4 L/mol.
Finally, multiply the number of moles of hydrogen gas by the molar volume at STP to get the volume of hydrogen gas:
Volume (in liters) = (3 * (moles of methane)) * 22.4 L/mol
Plug in the values for the moles of methane calculated earlier, and calculate the volume of hydrogen gas.