17.5 mL of oxygen gas were collected at room temperature (22oC) and 100.2 kPa of atmospheric pressure.
a) How many moles of oxygen gas were produced?
b) What is the molar volume of the oxygen gas at the conditions in the laboratory?
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A fixed mass of H2(g)occupies 450cm at 28.c calcalate it volume at 43.c what the pressure remain constant
To find the number of moles of oxygen gas produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in Pascal (Pa)
V = volume in cubic meters (m^3)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (K)
First, let's convert the given values to the correct units:
Volume: 17.5 mL = 17.5 / 1000 = 0.0175 L = 0.0175 / 1000 = 0.0175 m^3
Pressure: 100.2 kPa = 100.2 × 1000 = 100,200 Pa
Temperature: 22°C = 22 + 273.15 = 295.15 K
Substituting these values into the ideal gas law equation, we have:
(100200) × (0.0175) = n × (8.314) × (295.15)
Simplifying the equation:
1753.5 = 2449.5031n
Dividing both sides by 2449.5031, we find:
n = 1753.5 / 2449.5031 ≈ 0.715 moles
Therefore, approximately 0.715 moles of oxygen gas were produced.
Now, let's find the molar volume of the oxygen gas at the given conditions. The molar volume is defined as the volume occupied by one mole of a gas.
To find the molar volume, we can rearrange the ideal gas law equation to solve for V:
V = (nRT) / P
Using the same values as before, we can calculate:
V = (0.715) × (8.314) × (295.15) / 100200
Simplifying the equation:
V ≈ 0.0186 m^3
Therefore, the molar volume of the oxygen gas at the given conditions is approximately 0.0186 cubic meters (m^3) per mole.