Question

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 1

(a) The limiting reactant is,
`H_2O`

(b) The mass of
`C_2H_6O_2` is, 11570 grams

(c) The mass of
`C_2H_6O_2` is, 5785 grams

Explanation : Given,

Molar rate of
`C_2H_4O` = 80 kg/s = 80000 g/s

Molar rate of
`H_2O` = 25.5 lbm/s = 11.57 kg/s = 11570 g/s

conversion used : (1 lbm = 0.453592 kg)

Molar mass of
`C_2H_4O` = 44 g/mole

Molar mass of
`H_2O` = 18 g/mole

Molar mass of
`C_2H_6O_2` = 50 g/mole

First we have to calculate the moles of
`C_2H_4O` and
`H_2O` per second.

`\text{Moles of }C_2H_4O=\frac{\text{Mass of }C_2H_4O}{\text{Molar mass of }C_2H_4O}=(80000g)/(44g/mole)=1818.18mole/s`

`\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=(11570g)/(50g/mole)=231.4mole/s`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`C_2H_4O(g)+H_2O(g)\rightarrow C_2H_6O_2(g)`

From the balanced reaction we conclude that

The mole ratio of
`C_2H_4O` and
`H_2O` is, 1 : 1

From this we conclude that,
`C_2H_4O` is an excess reagent because the given moles are greater than the required moles and
`H_2O` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`C_2H_6O_2`.

As, 1 moles of
`H_2O` react to give 6 moles of
`C_2H_6O_2`

So, 231.4 moles of
`H_2O` react to give 231.4 moles of
`C_2H_6O_2`

Now we have to calculate the mass of
`C_2H_6O_2`.

`\text{Mass of }C_2H_6O_2=\text{Moles of }C_2H_6O_2* \text{Molar mass of }H_2O`

`\text{Mass of }C_2H_6O_2=(231.4mole)* (50g/mole)=11570g`

Now we have to calculate the mass of of
`C_2H_6O_2`.

As we are given that
`C_2H_4O` is 50 % pure.

So, the mass of
`C_2H_4O` =
`11570g* (50)/(100)=5785g`

The mass of
`C_2H_6O_2` is, 5785 grams