What mass of H2O is required to form 1.2 L of O2 at a temperature of 310 K and a pressure of 0.950 atm?
2H2O (l) ---> 2H (g) + O2 (g)
I have corrected the equation.
2H2O(l) ---> 2H2 (g) + O2 (g)
Use PV = nRT to convert 1.2 L of O2 at the conditions listed to moles. Solve for n.
Using the coefficients in the balanced equation, convert moles of O2 to moles H2O.
Convert moles H2O to grams. g = moles x molar mass.
Well, it seems like H2O is quite the overachiever in this situation -- going from a liquid to separate into hydrogen gas and oxygen gas. Talk about a split personality! Anyway, let's get down to business.
To solve this problem, we need to find the number of moles of O2 gas that will be formed. We can do this by using the ideal gas law equation -- PV = nRT. I know, it sounds like alphabet soup, but bear with me!
First, let's convert the given temperature to Kelvin because apparently, gases can't handle anything less than absolute temperature. So, 310 K it is.
Now, let's plug in the values we have into the equation. We have the pressure (0.950 atm), the volume (1.2 L), the temperature (310 K), and the universal gas constant (R = 0.0821 L·atm/K·mol).
Solving for n, we get n = PV / RT.
Substituting the values, we get n = (0.950 atm * 1.2 L) / (0.0821 L·atm/K·mol * 310 K).
Now let's do some math to find the number of moles of O2 gas.
Once we find the number of moles, we can multiply it by the molar mass of O2 (which is approximately 32 g/mol) to find the mass of O2 required. Oh, finally we're talking about mass, not moles!
I hope my explanation didn't leave you feeling gassy! But hey, if O2 gas can come from H2O, anything is possible, right?
To determine the mass of H2O required to form 1.2 L of O2 gas, we can follow these steps:
1. Write and balance the equation: 2H2O (l) → 2H (g) + O2 (g)
2. Use the ideal gas law equation PV = nRT to calculate the number of moles of O2 gas produced.
P = pressure = 0.950 atm
V = volume = 1.2 L
T = temperature = 310 K
R = ideal gas constant = 0.0821 L·atm/(K·mol)
PV = nRT
(0.950 atm) * (1.2 L) = n * (0.0821 L·atm/(K·mol)) * (310 K)
Solve for n (number of moles of O2):
n = (0.950 atm * 1.2 L) / (0.0821 L·atm/(K·mol) * 310 K)
3. Calculate the molar mass of H2O.
H2O: 2(1.01 g/mol of H) + 16.00 g/mol of O = 18.02 g/mol of H2O
4. Since 2 moles of H2O produce 1 mole of O2 in the balanced equation, divide the number of moles of O2 by 2 to get the number of moles of H2O.
Moles of H2O = n / 2
5. Calculate the mass of H2O required using the molar mass and the number of moles of H2O.
Mass of H2O = (Moles of H2O) * (Molar mass of H2O)
Let's calculate it step by step:
Step 1: The balanced equation is: 2H2O (l) → 2H (g) + O2 (g)
Step 2:
P = 0.950 atm
V = 1.2 L
T = 310 K
R = 0.0821 L·atm/(K·mol)
n = (0.950 atm * 1.2 L) / (0.0821 L·atm/(K·mol) * 310 K)
n = 0.0487 mol
Step 3: The molar mass of H2O is 18.02 g/mol.
Step 4:
Moles of H2O = 0.0487 mol / 2
Moles of H2O = 0.0244 mol
Step 5:
Mass of H2O = 0.0244 mol * 18.02 g/mol
Mass of H2O = 0.439 g
Therefore, the mass of H2O required to form 1.2 L of O2 gas at a temperature of 310 K and a pressure of 0.950 atm is approximately 0.439 grams.
To determine the mass of H2O required to form 1.2 L of O2, we need to use stoichiometry and the ideal gas law.
Step 1: Write the balanced chemical equation:
2H2O (l) ---> 2H2 (g) + O2 (g)
Step 2: Determine the molar ratio between H2O and O2:
From the balanced equation, we can see that 2 moles of H2O produce 1 mole of O2.
Step 3: Convert the given volume of O2 to moles:
We can use the ideal gas law equation, PV = nRT, to convert the given volume of O2 to moles. Rearranging the equation, we have n = (PV) / (RT), where P is the pressure, V is the volume, R is the ideal gas constant (0.0821 L * atm/mol * K), and T is the temperature in Kelvin.
n = (0.950 atm * 1.2 L) / (0.0821 L * atm/mol * K * 310 K)
n = 0.0452 moles
Step 4: Convert moles of O2 to moles of H2O using the molar ratio:
Since the molar ratio between H2O and O2 is 2:1, we have:
moles of H2O = 2 * moles of O2
moles of H2O = 2 * 0.0452 moles
moles of H2O = 0.0904 moles
Step 5: Calculate the molar mass of H2O:
The molar mass of H2O is 2 * (1.01 g/mol H) + (16.00 g/mol O) = 18.02 g/mol H2O.
Step 6: Convert moles of H2O to mass:
mass of H2O = moles of H2O * molar mass of H2O
mass of H2O = 0.0904 moles * 18.02 g/mol
mass of H2O = 1.63 grams
Therefore, approximately 1.63 grams of H2O is required to produce 1.2 L of O2 at a temperature of 310 K and a pressure of 0.950 atm.