Need guidance on how to solve this one
How many liters of water vapor can be produced if 108 grams of methane gas are conbusted at 312K and 98 atm. CH4 + 202 = C02 + 2H20
Well, let's see if we can bring some light-heartedness to this combustion conundrum!
To solve this question, we'll need to use stoichiometry, which is a fancy word for "making mole ratios dance".
First, we need to figure out how many moles of methane we have. Since methane (CH4) has a molar mass of 16 grams per mole (1 carbon atom at 12 atomic mass units + 4 hydrogen atoms at 1 atomic mass unit each), we divide 108 grams by 16 grams per mole to find that we have 6.75 moles of methane.
Now, let's peek at our balanced equation: 1 mole of methane produces 2 moles of water vapor. So if we have 6.75 moles of methane, we will have 6.75 times 2 (13.5) moles of water vapor.
But we're not searching for moles here; we want liters of water vapor. To get liters, we'll need to use the ideal gas law. At 312K and 98 atm, the ideal gas law states:
PV = nRT
Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin.
Since we're given the pressure (98 atm) and temperature (312K), we can solve for the number of liters of water vapor using the equation: V = nRT/P.
Substituting in the values, we find V = (13.5 moles * 0.0821 L·atm/(K·mol) * 312K) / 98 atm. After some math whizzing, we get V ≈ 43.04 liters.
So, approximately 43.04 liters of water vapor can be produced when 108 grams of methane gas are combusted at 312K and 98 atm.
And just like that, we've turned a combustion question into a heated dance of mole ratios and gas laws! Hope that made you smile a bit!
To determine how many liters of water vapor can be produced when 108 grams of methane gas (CH4) is combusted at 312K and 98 atm, we need to follow these steps:
Step 1: Convert the mass of methane gas (CH4) into moles.
To do this, we'll use the molar mass of methane (16.04 g/mol for carbon + 4 * 1.0078 g/mol for hydrogen).
Molar mass of CH4 = 12.01 g/mol (carbon) + 4 * 1.0078 g/mol (hydrogen) = 16.04 g/mol
Number of moles of CH4 = Mass of CH4 / Molar mass of CH4
Number of moles of CH4 = 108 g / 16.04 g/mol ≈ 6.729 mol
Step 2: Determine the molar ratio of water vapor (H2O) to methane (CH4) in the balanced chemical equation.
From the balanced chemical equation: CH4 + 2O2 = CO2 + 2H2O, we can see that for every 1 mole of methane (CH4), 2 moles of water vapor (H2O) are produced.
Step 3: Calculate the number of moles of water vapor produced.
Number of moles of H2O = 2 moles of H2O / 1 mole of CH4 * Number of moles of CH4
Number of moles of H2O = 2 * 6.729 mol ≈ 13.458 mol
Step 4: Use the ideal gas law to calculate the volume of water vapor.
The ideal gas law is expressed as PV = nRT, where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Rearranging the ideal gas law equation, we get V = (nRT) / P.
Since we are given the temperature T (312K) and pressure P (98 atm), we can plug in these values to calculate the volume V.
Volume of H2O (water vapor) = (Number of moles of H2O * R * T) / P
Volume of H2O = (13.458 mol * 0.0821 L·atm/(mol·K) * 312K) / 98 atm ≈ 43.544 L
Therefore, approximately 43.544 liters of water vapor can be produced when 108 grams of methane gas (CH4) is combusted at 312K and 98 atm.
To solve this question, we will use the stoichiometry of the combustion reaction and the ideal gas law.
Step 1: Find the moles of methane (CH4)
To find the moles of methane, we will use the molar mass of methane (CH4), which is approximately 16 grams per mole.
Moles of CH4 = Mass of CH4 / Molar mass of CH4
Moles of CH4 = 108 g / 16 g/mol
Moles of CH4 = 6.75 mol
Step 2: Use the stoichiometry of the combustion reaction
From the balanced combustion equation, we know that 1 mole of methane (CH4) produces 2 moles of water (H2O).
Moles of H2O produced = Moles of CH4 × 2
Moles of H2O produced = 6.75 mol × 2
Moles of H2O produced = 13.5 mol
Step 3: Convert moles of water to liters of water vapor
To convert moles of water to liters of water vapor, we will use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Rearranging the ideal gas law to solve for volume (V), we have:
V = nRT / P
Let's substitute the values and solve:
V = (13.5 mol) × (0.0821 L·atm/(mol·K)) × (312 K) / (98 atm)
V ≈ 35.95 L
Therefore, approximately 35.95 liters of water vapor can be produced if 108 grams of methane gas are combusted at 312 K and 98 atm.