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

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29-10-2022
Chemistry

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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

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NyimaU320458 NyimaU320458 · Answer 1 · 2022-10-29T21:52:18+02:00

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.

[tex]\begin{gathered} molC_2H_4=givengC_2H_4\times\frac{1molC_2H_4}{MolarMass,gC_2H_4} \\ molC_2H_4=81.7gC_2H_4\times\frac{1molC_2H_4}{28.05gC_2H_4}=2.91molC_2H_4 \end{gathered}[/tex]

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

[tex]\begin{gathered} molC_2H_4O=2.91molC_2H_4\times\frac{2molC_2H_4O}{2molC_2H_4} \\ molC_2H_4O=2.91molC_2H_4O \end{gathered}[/tex]

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:

[tex]\begin{gathered} molC_2H_4O=givengC_2H_4O\times\frac{MolarMass,gC_2H_4O}{1molC_2H_4O} \\ molC_2H_4O=2.91gC_2H_4O\times\frac{44.05gC_2H_4O}{1molC_2H_4O}=128.3gC_2H_4O \end{gathered}[/tex]

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