Question

How many litres of fluorine gas at stp can be collected from the decomposition of 90.7 g of AuF3

Answer 1

Answer: 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of
`AuF_3`

Step-by-step explanation:

The balanced decomposition reaction is shown as

`2AuF_3\rightarrow 2Au+3F_2`

moles of
`AuF_3=\frac{\text {given mass}}{\text {Molar mass}}=(90.7g)/(254g/mol)=0.36moles`

According to stoichiometry:

2 moles of
`AuF_3` gives = 3 moles of flourine gas

Thus 0.36 moles of
`AuF_3` gives =
`(3)/(2)* 0.36=0.54moles` of flourine gas

Using ideal gas equation :

`PV=nRT`

P = pressure of gas = 1 atm ( at STP)

V = Volume of gas = ?

n = moles of gas = 0.54

R = gas constant = 0.0821 L atm/Kmol

T = temperature = 273 K ( at STP)

Putting the values we get :

`1atm* V=0.54mol* 0.0821Latm/Kmol* 273K`

`V=12L`

Thus 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of
`AuF_3`