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Notice that in the last problem you calculated the grams of oxygen needed. what would the volume of this oxygen be at stp? report your answer in liters to the nearest whole number.

Sagot :

The volume of oxygen required at STP is 7 L.

STP is a pressure of 100 kPa and a temperature of 273.15 K. Or, 1 mole of any ideal gas occupies 22.7 L at STP.

Therefore, we need to find out the number of moles of oxygen required in the reaction.

Using the molar mass of water,

10.5 g × 1 mole of water/ 18.02 g = 0.577 moles of water.

1: 2 is the ratio between oxygen and water in a water molecule, which means 1 mole of oxygen produces 2 moles of water.

0.577 moles of H₂O × 1 mole O₂ / 2 moles of H₂O = 0.2885 moles O₂.

So, if 22.7 L at STP is occupied by 1 mole, then

0.2885  × 22.7 L / 1 mole = 6.548 L

Rounded to the nearest whole number, we get 7 L.

Hence, the volume of oxygen at STP is 7L.

To learn more about volume and STP, visit: https://brainly.com/question/15469566

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