What is the volume of 0.80 grams of o2 gas at stp?

The volume occupied by at STP is 0.56 L.

Explanation:

To calculate the number of moles, we use the equation:

We are given:

Given mass of oxygen = 0.80 g

Molar mass of oxygen = 32 g/mol

Putting values in above equation, we get:

At STP:

1 mole of a gas occupies 22.4 L of volume.

So, 0.025 moles of oxygen gas will occupy = of volume.

Hence, the volume occupied by at STP is 0.56 L.

Answer : The volume of oxygen gas at STP is, 0.56 L or 560 ml

Explanation : Given,

Mass of oxygen gas = 0.80 g

Molar mass of oxygen gas = 32 g/mole

First we have to calculate the moles of oxygen gas.

Now we have to calculate the volume of oxygen gas.

As we know that at STP,

1 mole of oxygen gas contains 22.4 L volume if oxygen gas

So, 0.025 mole of oxygen gas contains volume if oxygen gas

Conversion : 1 L = 1000 ml

Therefore, the volume of oxygen gas at STP is, 0.56 L or 560 ml

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it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2