Question

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

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 = (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. =)