Ethanol, C 2 H 5 OH, is an ingredient in alcoholic beverages and an octane boost

Question

Ethanol, C2H5OH, is an ingredient in alcoholic beverages and an octane booster in gasoline. It's produced by the fermentation of glucose by the following reaction:

C6H12O6(aq) 2 C2H5OH(aq) + 2 CO2(g)

If the change in the concentration of ethanol over time is 7.622 x 10-4 M/s, how many moles of ethanol are produced in exactly 24 hours in a total solution volume is 10.00 L?

The rate constant for the zeroth-order decomposition of NH3 on a platinum surface at a given temperature is 4.38 x10-6 M/s. If the initial concentration of NH3 is 2.74x10-2 M, what is the concentration of NH3 after 30.0 minutes?  

Answer in decimal format.

Explanation / Answer

Given Rate = 7.622e-4 M/s

Time = 24 hours = 24 x 60 x 60 = 86400sec

So concentration after 86400sec = 7.622e-4 x 86400 = 65.85M

So, moles = molarity x volume = 65.85 x 10 L = 658.5moles

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