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What volume, in liters, does 128 grams of O2 occupy at STP?89.6 L22.4 L67.2 L44.8 L

User Shaunc
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1 Answer

5 votes

Answer:

89.6 L

Step-by-step explanation:

The volume of a gas at STP (Standard Temperature and Pressure) is 22.4 L per mole of gas. To determine the volume of 128 grams of O2, we need to convert the mass of O2 to moles of O2:

First, we need to determine the number of moles of O2 in 128 grams of O2 using the molar mass of O2:

molar mass of O2 = 2 x atomic mass of O = 2 x 16.00 g/mol = 32.00 g/mol

moles of O2 = mass of O2 / molar mass of O2

moles of O2 = 128 g / 32.00 g/mol = 4.00 mol

Now that we have the number of moles of O2, we can use the volume at STP for one mole of gas to calculate the total volume of the sample:

volume of O2 = number of moles of O2 x volume at STP per mole of gas

volume of O2 = 4.00 mol x 22.4 L/mol = 89.6 L

Therefore, 128 grams of O2 at STP occupies a volume of 89.6 L.

User Paddington
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