How many liters of oxygen (at STP) are required to form 10.5 g of #H_2O#? #2H_(2(g)) + O_(2(g)) -> 2H_2O_((g))#

To form that much water at STP, 6.55 L of oxygen would be required.

STP conditions, which are described as a temperature of 273.15 K and a pressure of 100 kPa, should cause you to consider a gas's molar volume right away.

Specifically, regarding the fact that, at STP, one mole of any ideal gas has an exact volume of 22.7 L.

This means that the amount of oxygen that really needs to be determined is just the number of moles that participated in the reaction. Once the number of moles of oxygen required is known, the volume of oxygen can be calculated using the molar volume of a gas at STP.

Calculate how many moles of water the reaction would produce using the molar mass of water.

Examine the balanced chemical equation now.

Determine how many moles of oxygen there are by

Consequently, if 1 mole takes up 22.7 L at STP, then

The response, rounded to three sig figs, is

SIDE NOTE: Generally speaking, you will have to apply the previous definition of STP, which calls for a temperature of 273.15 K and a pressure of 1 atm. One mole of any ideal gas would take up 22.4 L in these circumstances.

The molar volume of 22.7 L is implied by the current definition of STP; however, if your teacher or instructor requests that you use the previous value, just redo the final calculation using 22.4 L rather than 22.7 L.

Using stoichiometry, we can calculate the amount of oxygen required to form 10.5 g of H2O.

1 mole of H2O = 18.015 g 2 moles of H2O = 36.03 g

According to the balanced chemical equation, 2 moles of H2O are produced from 1 mole of O2.

So, to produce 36.03 g of H2O, we need 1 mole of O2.

Since 1 mole of any gas at STP occupies 22.4 liters, we need 22.4 liters of O2 to produce 36.03 g of H2O.

Now, to find the amount of O2 required to produce 10.5 g of H2O:

(10.5 g H2O) x (22.4 L O2 / 36.03 g H2O) = 6.55 L O2