Chemical reactions and stoichiometry q2. ethylene glycol (c2hs02) can be produced from the reaction of ethylene oxide (c2h40) and water as shown below. 2h40(g) t 20(g) c2h602(0) if 80 kg/s of c2h40 is mixed with 25.5 lbm/s of water, a) what is the limiting reactant? b) how much ethylene glycol is produced when the reaction is complete? c) how much ethylene glycol is produced if c2h40 is 50% pure

Answer :

(a) The limiting reactant is,

(b) The mass of is, 11570 grams

(c) The mass of is, 5785 grams

Explanation : Given,

Molar rate of = 80 kg/s = 80000 g/s

Molar rate of = 25.5 lbm/s = 11.57 kg/s = 11570 g/s

conversion used : (1 lbm = 0.453592 kg)

Molar mass of = 44 g/mole

Molar mass of = 18 g/mole

Molar mass of = 50 g/mole

First we have to calculate the moles of and per second.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

The mole ratio of and  is, 1 : 1

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of .

As, 1 moles of react to give 6 moles of

So, 231.4 moles of react to give 231.4 moles of

Now we have to calculate the mass of .

Now we have to calculate the mass of of .

As we are given that is 50 % pure.

So, the mass of =

The mass of is, 5785 grams

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