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What is the volume of oxygen O2 found in 41.1 grams of oxygen at STP?

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User Ravitheja
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Answer:

2.637 L

Step-by-step explanation:

The volume of oxygen O2 found in 41.1 grams of oxygen at STP can be calculated using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles of oxygen, R is the ideal gas constant, and T is the temperature.

Using the molar mass of oxygen, which is 32 g/mol, we can calculate the number of moles of oxygen in 41.1 g of oxygen:

n = 41.1 g/ 32 g/mol = 1.28125 moles

At STP, the pressure is 101.325 kPa and the temperature is 273.15 K. Plugging these values into the ideal gas law equation gives us:

V = (1.28125 mol)(0.08206 L•atm/mol•K)(273.15 K) / (101.325 kPa) = 2.637 L

User JudgeProphet
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