Find the total volume of the mixture of hydrogen gas and oxygen gas obtained from the reaction of 130.0 gram of water at 25.0 c and 1.00 atm pressure?

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19-08-2022
Chemistry

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Find the total volume of the mixture of hydrogen gas and oxygen gas obtained from the reaction of 130.0 gram of water at 25.0 c and 1.00 atm pressure?

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pankajpal6971 pankajpal6971 · Answer 1 · 2022-08-24T13:06:35+02:00

The volume of the mixture of hydrogen gas and oxygen gas obtained from the reaction of 130.0 gram of water at 25.0 c and 1.00 atm pressure is 176.15 L.

Calculation of number of moles of H2O

Moles is defined as the ratio of given mass to molar mass of compound.

Given,

mass of H2O = 130 g

molar mass of H2O = 18 g

Moles = given mass/ molar mass

= 130/ 18

= 7.2 mol

Calculation of volume of H2O

By using ideal gas equation,

PV = nRT

where,

P is the pressure of H2O

V is the volume of H2O

R is the gas constant.

T is the temperature of H2O

Given,

P = 1 atm

T = 25 + 273 = 298K

n = 7.2 mol

R = 0.0821 atm L/ mol K

1 × V = 7.2 × 0.0821× 298

V = 7.2 × 0.0821× 298

V = 176.15 L

Thus, we calculated that the volume of the mixture of hydrogen gas and oxygen gas obtained from the reaction of 130.0 gram of water at 25.0 c and 1.00 atm pressure is 176.15 L.

learn more about ideal gas equation:

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