Question

At one time, acetaldehyde was prepared industriallyby the reaction of ethylene with air in the presence ofa copper catalyst. How many grams of acetaldehydecan be prepared from 81.7 g of ethylene?2C₂H4(g) + O₂(g)Ethylene =>2C₂H4O(g)Acetaldehyde

Answer 1

We must first find the moles corresponding to 81.7 grams of ethylene. To do this, we divide the grams by the molar mass of ethylene: 28.05g/mol.

`\begin{gathered} molC_2H_4=givengC_2H_4*(1molC_2H_4)/(MolarMass,gC_2H_4) \\ molC_2H_4=81.7gC_2H_4*(1molC_2H_4)/(28.05gC_2H_4)=2.91molC_2H_4 \end{gathered}`

Now, by stoichiometry, we see that the ratio of acetaldehyde to ethylene is 2/2. So, the moles of C2H4O will be:

`\begin{gathered} molC_2H_4O=2.91molC_2H_4*(2molC_2H_4O)/(2molC_2H_4) \\ molC_2H_4O=2.91molC_2H_4O \end{gathered}`

Now, to find the grams of C2H4O we will need the molar mass of C2H4O. The molar mass of C2H4O is 44.05g/mol. The grams of C2H4O will be:

`\begin{gathered} molC_2H_4O=givengC_2H_4O*(MolarMass,gC_2H_4O)/(1molC_2H_4O) \\ molC_2H_4O=2.91gC_2H_4O*(44.05gC_2H_4O)/(1molC_2H_4O)=128.3gC_2H_4O \end{gathered}`

Answer: The grams of acetaldehyde that can be prepared are 128.3 grams