Question

2. Consider the following reaction: 2A1 () + 6HCl(aq) + 2AICI 3(aq) + 3H 28) A 0.040 mole piece of aluminum reacted completely in 20 s. The rate of formation of hydrogen gas is: A. 0.0013 mol/s B. 0.0020 mol/s C. 0.0030 mol/s D. 0.0060 mol/s

Answer 1

Rate of consumption of Al = -∆[Al]/∆t 0.040/20 = 0.002 mol/s

Now, rate = -1/2 ∆[Al]/∆t = 1/3 ∆[H2]/∆t

Rate of formation of H2 = ∆[H2]/∆t = -3/2 ∆[Al]/∆t

= 3×0.002/2

= 0.003 mol/s

Answer is c)

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