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wlad13 [49]
3 years ago
15

Assuming 1 mol of Fe3+ and 2 mol of SCN- were allowed to react and reach equilibrium. 0.5 mol of product was formed. The total v

olume at equilibrium was 1 L. How much Fe3+ remained at equilibrium? (in mol) (1 point) Your Answer: How much SCN- remained at equilibrium? (in mol) (1 point) Your Answer: What’s the equilibrium constant, Kc? (1 point)
Chemistry
1 answer:
zaharov [31]3 years ago
3 0

Answer:

a. 0.5 mol

b. 1.5 mol

c. 0.67

Explanation:

Fe3+ + SCN- -----> [FeSCN]2+

a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium

b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium

c. KC =  0.5/(0.5*1.5) =  0.67

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