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 $AuF_3$

Explanation:

The balanced decomposition reaction is shown as

$2AuF_3\rightarrow 2Au+3F_2$

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

According to stoichiometry:

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

Thus 0.36 moles of $AuF_3$ gives = $\frac{3}{2}\times 0.36=0.54moles$ 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 :

$1atm\times V=0.54mol\times 0.0821Latm/Kmol\times 273K$

$V=12L$

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