Respuesta :
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.
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