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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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Answer:

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

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