Respuesta :

McChuck
The conversion factor for volume at STP is [tex]\frac {1mol}{22.4L}[/tex] or \frac {22.4L}{1mol}. Since we want volume, we would use \frac {22.4L}{1mol}. We conclude with the following calculations:

[tex]2molCl_{2}*\frac {22.4LCl_{2}}{1molCl_{2}} = 44.8L Cl_{2}[/tex]

The answer is 44.8L Cl2
JannetPalos

Answer:

Volume of 2 mol of chlorine gas at STP is 44.8 L

Explanation:

Let us assume chlorine gas behaves ideally.

According to law of ideal gas-      PV=nRT

where P is pressure of gas, V is volume of gas, n is number of moles of gas, R is gas constant and T is temperature in Kelvin.

At STP, P is 1 atm and T is 273 K

Here n=2 and R=0.082 L.atm/mol.K

So [tex]V=\frac{nRT}{P}=\frac{2\times 0.082\times 273}{1}L=44.8 L[/tex]

Hence volume of chlorine gas at STP is 44.8 L

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