Question #193635

How many grams of chlorine gas, Cl2, are in 3.12 L at -14 ° C and 253 kPa pressure?

Expert's answer

P = 253 kPa

V = 3.12 L

R = 8.314 kPa L K^{-1} mol^{-1}

T = -14°C = 259 K

**Solution:**

The Ideal Gas equation can be used.

The Ideal Gas equation can be expressed as: PV = nRT

To find the moles of chlorine gas, solve the equation for n:

n(Cl_{2}) = PV / RT

n(Cl_{2}) = (253 kPa × 3.12 L) / (8.314 kPa L K^{-1} mol^{-1} × 259 K) = 0.3666 mol

Moles of Cl_{2} = 0.3666 mol

The molar mass of Cl_{2} is 70.906 g/mol.

Therefore,

(0.3666 mol Cl_{2}) × (70.906 g Cl_{2} / 1 mol Cl_{2}) = 25.994 g Cl_{2} = **26 g Cl**_{2}

**Answer: 26 grams of chlorine gas (Cl**_{2}**).**

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