Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the mixture of NO and O2 shown in the accompanying diagram. The blue spheres represent N, and the red ones represent O. (c) If the actual yield of the reaction was 75% instead of 100%, how many molecules of each kind would be present after the reaction was over?

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Hi everyone here we have a question telling us that the reaction between sulfur dioxide and oxygen produces sulfur trioxide with the yield of 55% for the mixture shown below, determine the number of molecules for each compound after the reaction. So let's see how many molecules of software dioxide we have. We have 1234567. And let's see how many oxygen. We have.  aN:aN:000NaN 123456. And now let's write our balanced equation. So we have sulfur dioxide plus oxygen produces sulfur trioxide, and we have four oxygen's on the left and three on the right. So we're going to put a two in front of our sulfur dioxide and a two in front of our software trioxide. So now we need to decide the limiting re agent. So we have seven moles of sulfur dioxide times two moles a sulfur trioxide over two moles A sulfur dioxide. And our moles of software dioxide will cancel out. And that will equal seven moles of sulfur trioxide. And then we have six moles of oxygen times two moles of sulfur trioxide over one mole Of oxygen, and our moles of oxygen will cancel out. And that will give us 12 moles of sulfur trioxide. So sulfur dioxide is our limiting re agent because it produces the least amount of sulfur trioxide. So now we need to calculate our percent yield. So we have seven moles of software trioxide times 55 Over because it was 55% and that equals 3.85 moles of sulfur trioxide. And now we need to calculate our access our access So to produce 3.85 moles of sulfur trioxide. We're going to multiply by two moles of sulfur dioxide over two moles. A sulfur trioxide And our moles of sulfur trioxide are going to cancel out and that's going to give us 3.85 moles of sulfur dioxide consumed. And Next we're going to do the same with oxygen. So 3.85 moles That's sulfur trioxide times one mole of oxygen over two moles of sulfur trioxide Equals 1.925 moles of oxygen consumed. Our moles of sulfur dioxide and excess equals seven moles of sulfur dioxide, initial -3.85 moles of sulfur dioxide consumed equals three 0.15 moles of sulfur dioxide. Our moles of 02 in excess equals six moles of oxygen, Initial -1.925 moles of oxygen consumed Equals 4.075 moles of oxygen. So these are our final answers. Thank you for watching. Bye!

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Key Concepts

Stoichiometry

Percent Yield

Molecular Representation