Question 11 What/mass of oxygen is needed to fill a 585 L tank at STP?
0.417 kg
0.835 kg
0.765 kg

0.835 Kg

Explanation:

We'll begin by calculating the number of mole of oxygen that will fill 585 L tank at STP. This can be obtained as follow:

1 mole of oxygen (O2) occupies 22.4 L at STP.

Therefore, Xmol of oxygen will occupy 585 L tank at STP i.e

Xmol of oxygen = 585/22.4

Xmol of oxygen = 26.12 moles

Next, we shall determine the mass of 26.12 moles of oxygen. This can be obtained as follow:

Mole of O2 = 26.12 moles

Molar mass of O2 = 2 × 16 = 32 g/mol

Mass of O2 =?

Mole = mass /Molar mass

26.12 = mass of O2 /32

Cross multiply

Mass of O2 = 26.12 × 32

Mass of O2 = 835.84 g

Finally, we shall convert 835.84 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

835.84 g = 835.84 g × 1 Kg / 1000 g

835.84 g = 0.835 Kg

Thus, 835.84 g is equivalent to 0.835 Kg.

Therefore, the mass of oxygen needed to fill the 585 L tank at STP is 0.835 Kg

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