How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26°C? Show all work used to solve this problem.
Use PV = nRT and solve for n. Remember that T must be in kelvin.
Then n = grams/molar mass. Solve for grams.
can someone solve this?(21.3 g Li/6.941 g Li/moleLi)*(1mole H2/1moleLi) * 22.4 L at STP * ( temp correction 297/273)* (pressure correction 1/1.4) = your answer in L of H2
maybe this is better
If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmospheres? Show all of the work used to solve this problem.
To solve this problem, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, let's convert the given values into the appropriate units:
Pressure: 0.98 atmospheres
Volume: 10.2 liters
Temperature: 26°C = 26 + 273 = 299 Kelvin
Substituting the values into the equation, we get:
(0.98 atm) * (10.2 L) = n * (0.0821 L·atm/mol·K) * (299 K)
Simplifying the equation, we have:
10.096 atm·L = n * 24.5119 L·atm/mol
Now, we can solve for n (number of moles):
n = (10.096 atm·L) / (24.5119 L·atm/mol)
n ≈ 0.412 mol
Next, we need to convert the number of moles to grams using the molar mass of water (H2O):
Molar mass of H2O: 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
Mass = n * molar mass
Mass = (0.412 mol) * (18.016 g/mol)
Mass ≈ 7.410 g
Therefore, there are approximately 7.410 grams of water vapor in the 10.2 liter sample at 0.98 atmospheres and 26°C.