Determine the volume (in L) of Cl2 (g) required to carry out the following reaction at 794 torr and 625°C using 15.0 g of Fe. The value of R = 0.0821 L atm mol-1 K-1. 2Fe (s) + 3Cl2 (g) = 2FeCl3 (s)

Question

Lorena Whitaker

Chemistry

2 November, 02:24

Determine the volume (in L) of Cl2 (g) required to carry out the following reaction at 794 torr and 625°C using 15.0 g of Fe. The value of R = 0.0821 L atm mol-1 K-1. 2Fe (s) + 3Cl2 (g) = 2FeCl3 (s)

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Seamus Daniels · Answer 1

28.5 L of Cl

Explanation:

From PV=nRT

P = 794torr or 1.04 atm

T = 625°C or 898K

R = 0.0821 L atm mol-1 K-1

n = 3 moles

V=?

V = nRT/P

V=3 * 0.0821 * 898/1.04

V = 212.7 L

From the balanced reaction equation

112g of iron reacted with 212.7L of Cl

15.0 g of iron will react with 15.0*212.7/112

= 28.5 L of Cl