Question

How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm?

Answer 1

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:
$P*V = n*R*T$

$1*18.5 = n*0.082*300$

$18.5 = 24.6n$

$24.6n = 18.5$

$n = \dfrac{18.5}{24.6}$

$\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark$

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

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I hope this helps. =)