PART 1
Given : 1) Mass of Iron = 0.558 g & 2) Mass of Iron + Mass of Oxygen = 0.718 g
Mass of Iron + Mass of Oxygen = 0.718 g
Mass of Oxygen = 0.718 g - mass of Iron = 0.718 g - 0.558 g = 0.160 g
Now, calculate no. of moles of Fe and O.
We have, No. of moles = Mass / Molar Mass
No. of moles of Fe = 0.558 g / (55.85 g /mol ) = 0.00999 mol Fe
No. of moles of O = 0.160 g / ( 16.00 g / mol ) = 0.0100 mol O
Ratio of number of moles of Fe : O is
0.00999 / 0.00999 = 1.00 mol Fe
0.0100 / 0.00999 = 1.00 mol O
In given oxide, Fe and O are present in the 1 : 1 ratio. Hence, empirical formula of oxide is FeO.
ANSWER : Empirical formula of oxide = FeO.
PART 2
Given : Total pressure of mixture of a gas = 925 torr
Partial pressure of O 2 gas = 425 torr
Partial pressure of Hegas = 75 torr
According to Dalton's law , the total pressure exerted by a mixture of non reacting gases is equal to the sum of partial pressures of individual gases.
Mathematically ,we can write P total = P 1 + P 2+ P 3 + .................
P total = P oxygen + P Helium + P Nitrogen
P Nitrogen = P total - ( P oxygen + P Helium )
P Nitrogen = 925 torr - ( 425 torr + 75 torr )
P Nitrogen = 925 torr - 500 torr
P Nitrogen = 425 torr
ANSWER : Partial pressure of nitrogen gas in the mixture = 425 torr
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