What volume (l) of fluorine gas is required to react with 2.31 g of calcium bromide to form calcium fluoride and bromine gas at 8.19 atm and 35 degrees celsius? (correct asnwer: 3.57x10^-3) but i got 3.57x10^-2

Hey there!:

Balanced equation:

F2(g) + CaBr2(s) ---> CaF2 + Br2

1 mol F2 = 1 mol CaBr2

Calculation:

2.31 g CaBr2 * (1 mol CaBr2 / 199.886 g CaBr2)  * (1 mol F2 / 1 mol CaBr2)

= 0.0115565 moles of F2:

Assuming F2 is an ideal  

gas at these conditions:

P*V = n*R*T

Solving for V:

V = (n *R* T) / P

where :

n = 0.0115565 moles

R = 0.08206 atm·L/mol·K

Temperature in K = 35 + 273.15 => 308.15 K

P = 8.19 atm

Substituting numbers into V = (n x R x T) / P:

V = 0.0115565 * 0.08206 * 308.15 / 8.19

V = 0.29222 / 8.19

V = 0.0357 or 3.57*10⁻² L

You are correct!!!

Hope that helps!