The combustion of liquid ethanol (C2H5OH) produces carbon
dioxide...

Question

The combustion of liquid ethanol (C2H5OH) produces carbon
dioxide and water. After 4.62 mL of ethanol (density =
0.789 g>mL) is allowed to burn in the presence of 15.55 g of oxygen
gas, 3.72 mL of water (density = 1.00 g>mL) is collected.
Determine the percent yield for the reaction. (Hint: Write a balanced equation
for the combustion of ethanol.)

Accepted Answer

Hello everyone today, we are being given a problem and asked to solve a 4% yield. The question reads when 24.6 g of C three H six or propane and 56.9 g of chlorine gas react. It produces an actual yield of 32.4 g of C three H five cl And then we have the chemical formula below and thankfully it's already balanced so we can go ahead and get started. Step one. We're going to want to find the limiting reacting. And so that means we're gonna take our given values for each reactant and convert them to moles of the product, which is C three H five cl. So we're gonna take 24.6 g of C three H 6. And we're gonna do the molar mass of we're gonna say one mole of the C three H six is equal to 42.8 g of C three H six. Now, since we want to convert to moles of C three H five cl we're gonna use the multiple ratio here. So we're gonna say one mole of C three H five cl Is produced from one mole of C three H six. And that is from our coefficients in front of each reactant and product. And so our units are going to cancel And we're left with 0.5846 moles of C three H five cl We're gonna do the same thing with chlorine gas. We're gonna say 56.9 g of cl two. We're going to convert them all. The molar mass. We're gonna say one mole of cl two is equal to 70.91 g of cl two. And then we're gonna do the multiple ratio. So we're going to say one mole of cl two produces one mole of c three H five cl. Our units cancel once again and we're left with 0.80-4 moles of C H five cl. Now the limiting reactant is whichever reactant produces the least amount of product and in this case .5846 moles is the least amount. So our limiting reactant, limiting reactant And the situation is C three H six. And so we're going to take how many moles we produce from that limiting reactant. In our second step we're going to find the theoretical yield, find theoretical yield. So we're gonna take 0.5846 moles of c three H five cl. And we're going to multiply it by its molar mass ratio. So we're going to say one mole of c three H five cl is equal to 76 53 g of c three H five cl. This gives us 44.74 g of c three H five cl When our units cancel out Our 3rd and final step is to find the actual yield. And we do that by doing the actual over the theoretical Times 100% because our answer has to be a percentage. So our actual yield is 32.40 g of c three h five cl. And our theoretical, what we just solved for is 44.74 g of c three h five cl. We multiply that by 100% and we get a final answer of 72.42%. I hope this helped, and until next time.

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