How many liters of hydrogen gas can be produced if 102 grams of methane gas (CH4) are combusted at 315 K and 1.2 atm? Show all of the work used to solve this problem.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g)
Without arrows you don't know the difference between reactants and products.
Here is a site that explains stoichiometry problems with a worked example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
You will need to find moles; use PV = nRT and use the conditions cited in the problem.
To solve this problem, we need to use the stoichiometry of the balanced chemical equation to convert grams of methane gas (CH4) to liters of hydrogen gas (H2).
Here are the steps to solve the problem:
Step 1: Convert grams of methane gas (CH4) to moles.
To convert grams to moles, we need to use the molar mass of methane, which is calculated by adding up the atomic masses of the atoms in the molecule.
The atomic masses:
C (carbon) = 12.01 g/mol
H (hydrogen) = 1.008 g/mol
The molar mass of methane (CH4):
(12.01 g/mol * 1 atom of C) + (1.008 g/mol * 4 atoms of H) = 16.04 g/mol
So, 102 grams of CH4 is equal to:
102 g / 16.04 g/mol = 6.36 moles of CH4
Step 2: Use the stoichiometry of the balanced equation to convert moles of CH4 to moles of H2.
From the balanced equation, we can see that for every 1 mole of CH4, we produce 2 moles of H2.
So, 6.36 moles of CH4 will produce:
6.36 moles CH4 * 2 moles H2 / 1 mole CH4 = 12.72 moles of H2
Step 3: Convert moles of H2 to liters of H2 gas.
To convert moles to liters, we need to use the ideal gas law equation:
PV = nRT
Where:
P is the pressure in atm
V is the volume in liters
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
Given:
Pressure (P) = 1.2 atm
Temperature (T) = 315 K
Number of moles of H2 (n) = 12.72 moles
Ideal gas constant (R) = 0.0821 L·atm/mol·K
Using PV = nRT, we can rearrange the equation to solve for V (volume):
V = (n * R * T) / P
Plugging in the values:
V = (12.72 moles * 0.0821 L·atm/mol·K * 315 K) / 1.2 atm
Calculating:
V = 259.813 liters
Therefore, the number of liters of hydrogen gas produced is approximately 259.813 liters.