How many moles of nacl will react completely with 18.5 l f2 gas at 300.0 k and 1.00 atm? f2(g) + 2nacl(s) yields cl2(g) + 2naf(s) 0.751 mol nacl 1.33 mol nacl 1.50 mol nacl 2.66 mol nacl

Hello!

Data:

P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)

Apply the data to the Clapeyron equation (ideal gas equation), see:












Note: If the feedback is to be considered, the closest response is 0.751 mol Nacl

_________________
_________________

I hope this helps. =)

Show an image of the data (˵¯͒ བ¯͒˵)