anitamosally715
contestada

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

Respuesta :

dexteright02
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:
[tex]P*V = n*R*T[/tex]

[tex]1*18.5 = n*0.082*300[/tex]

[tex]18.5 = 24.6n[/tex]

[tex]24.6n = 18.5[/tex]

[tex]n = \dfrac{18.5}{24.6} [/tex]

[tex]\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark[/tex]

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

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I hope this helps. =)


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