(16 pts) 10.0 g of iron metal was reacted with 15.0 g of chlorine gas to produce

Question

(16 pts) 10.0 g of iron metal was reacted with 15.0 g of chlorine gas to produce iron (III) chloride according to this balanced chemical reaction: 2 Fe(s) + 3 Cl. (g) 2 FeCl,(s) AH-_ 7990 kJ In the lab, the reaction yielded 19.74 g of product. a. Calculate the theoretical yield of iron (III) chloride product. b. Calculate the percent yield. c. Use the actual yield to estimate the amount of energy (in kJ) released by the reaction as conducted in the lab. d. How many liters of chlorine gas reacted at STP? e. Calculate the kinetic energy of chlorine gas at STP

Explanation / Answer

(a)

Moles of Fe = Mass/MW = 10/55.84 = 0.179

Moles of Cl2 = Mass/MW = 15/71 = 0.211

1 mole Fe requires 1.5 moles of Cl2 for complete reaction.

So, Cl2 is the limiting reagent because it is present in lesser amount than required.

So, theoretical yield = 0.211*(2/3) = 0.141 moles

Theoretical mass of product = Moles*MW = 0.141*162.2 = 22.87 g

(b)

Percent yield = (19.74/22.87)*100 = 86.31%

(c)

Amount of energy released by the reaction = 799*(86.31/100) = 689.61 kJ

(d)

Volume of Chlorine gas that reacted at STP = 22.4*0.211 = 4.73 L

Hope this helps !

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