Respuesta :
Answer: 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of [tex]AuF_3[/tex]
Explanation:
The balanced decomposition reaction is shown as
[tex]2AuF_3\rightarrow 2Au+3F_2[/tex]
moles of [tex]AuF_3=\frac{\text {given mass}}{\text {Molar mass}}=\frac{90.7g}{254g/mol}=0.36moles[/tex]
According to stoichiometry:
2 moles of [tex]AuF_3[/tex] gives = 3 moles of flourine gas
Thus 0.36 moles of [tex]AuF_3[/tex] gives = [tex]\frac{3}{2}\times 0.36=0.54moles[/tex] of flourine gas
Using ideal gas equation :
[tex]PV=nRT[/tex]
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 :
[tex]1atm\times V=0.54mol\times 0.0821Latm/Kmol\times 273K[/tex]
[tex]V=12L[/tex]
Thus 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of [tex]AuF_3[/tex]
