Answer to Question #154211 in General Chemistry for Sage Topaz

Question

Ethylene Glycol is used in automobile antifreeze and preparation of polyester fiber. Ethylene glycol is produced in the following reaction:

C₂H₂O  + H₂O →  C₂H₆O₂  

If 90 g of water are used completely in the reaction, how many moles of ethylene glycol are produced? How many grams is this equivalent to?

Answer 1

Q154211

Ethylene Glycol is used in automobile antifreeze and preparation of polyester fiber. Ethylene glycol is produced in the following reaction:

C₂H₂O  + H₂O → C₂H₆O₂  

If 90 g of water is used completely in the reaction, how many moles of ethylene glycol is produced? How many grams is this equivalent to?

Solution:

Ethylene glycol is prepared from ethylene (C₂H₄). The intermediate in the reaction is ethylene oxide(C₂H₄O ). Ethylene oxide reacts with water to form ethylene glycol.

C₂H₄O  + H₂O → C₂H₆O₂  

Maybe the reaction given in the question is incorrect and it is C₂H₄O instead of C₂H₂O.

Please check the reaction again.

We will consider C₂H₄O instead of C₂H₂O and solve the problem.

So the correct reaction is

C₂H₄O  + H₂O → C₂H₆O₂  

The mass of water used in the reaction is 90grams. Since all of the water is consumed in the reaction so we can consider that H₂O is the limiting reactant and the moles of

ethylene glycol formed can be calculated using the mass of water.

Step 1: Convert 90 grams of H₂O to moles.

molar mass of H₂O = 2 * atomic mass of H + 1 * atomic mass of O

= 2 * 1.00794g/mol + 1 * 15.999g/mol

= 2.01588g/mol + 15.999g/mol

= 18.015g/mol

Step 2: Using mole to mole ratio of H₂O and  C₂H₆O₂  from the reaction find the moles

of ethylene glycol formed from 4.996 mol H₂O.

The reaction is 1C₂H₄O  + 1H₂O → 1 C₂H₆O₂  

The mole to mole ratio of H₂O and  C₂H₆O₂  in this reaction is 1:1.

moles of C₂H₆O₂  = 4.996 mol H₂O * 1mol C₂H₆O₂  /1 mol H₂O

= 4.996 mol C₂H₆O₂.

in the given question 90 g is given in 1 significant figure, so our final answer must also be

in 1 significant figure.

Hence the moles of ethylene glycol formed is 5 moles.

Next, we convert 5 mol C₂H₆O₂ to grams using the molar mass of C₂H₆O₂.

molar mass of C₂H₆O₂

= 2 * atomic mass of C + 6 * atomic mass of H + 2 * atomic mass of O

= 2 * 12.011 g/mol + 6 * 1.00794 g/mol + 2 * 15.999 g/mol

= 24.022g/mol + 6.0476g/mol + 31.998g/mol

= 62.0676g/mol

Hence, moles of ethylene glycol produced in the given reaction is 5 moles and this is equivalent to 300 grams of ethylene glycol.