PLEASE GIVE ME SOME HELP

Question

For the following reaction, 5.08 grams of hydrogen gas are mixed with excess ethylene (C₂H₄) . The reaction yields 51.3 grams of ethane (C₂H₆) .

hydrogen ( g ) + ethylene (C₂H₄) ( g )  ethane (C₂H₆) ( g )

What is the theoretical yield of ethane (C2H6) ?	grams
What is the percent yield for this reaction ?	%

Answer 1

Hi Thomas!

Theoretical yield assumes that our reaction goes to completion, therefore our reactants are consumed 100%. To solve this question, we can break it into steps.

1. Balance the equation if it is not balanced. As it stands right now, the equation is balanced so all molecules in the reaction are in a 1:1:1 ratio.
2. Identify which reactant is our limiting one. The question states that ethylene is in excess, therefore we can assume that hydrogen gas is our limiting reactant.
3. Figure out the amount of product that our limiting reactant forms. The mentioned 51.3 grams ethane is the "actual" amount produced, meaning that if someone did this reaction in real life and got 51.3 grams of ethane. With that in mind, we can use dimensional analysis to convert our 5.08 grams of hydrogen gas (H2) to ethane.
4. First we convert grams to moles. 5.08 g H2 x (1 mol H2 / 2 g) = 2.54 moles of H2
5. Second we use the balanced chemical equation to determine the amount of ethane we would get with 2.54 moles of H2. 2.54 moles of H2 x (1 mol ethane / 1 mol H2) = 2.54 moles of ethane
6. Third we convert moles to grams using the molecular weight of ethane. Ethane has the molecular formula of C2H6 = 2 carbons + 6 hydrogens = (2 x 12 g/mol) + (6 x 1 g/mol) = 30 g/mol. With this in mind, we can convert moles to grams. 2.54 mol ethane x (30 g / 1 mol) = 76.2 grams of ethane. Therefore, 76.2 grams is our theoretical yield.

Percent yield is where we would take the mentioned 51.3 grams of ethane into account. Percent yield has the equation of actual yield / theoretical yield. Therefore, we can plug our two masses into the equation to get us an answer. 51.3 grams of ethane / 76.2 grams of ethane = 0.673 = 67.3% is our percent yield.

I hope this helps! Please let me know if you have questions regarding the solution.