Urea (CH₄N₂O) is a common fertilizer that is synthesized by the reaction of ammonia (NH₃) with carbon dioxide: 2 NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 136.4 kg of ammonia with 211.4 kg of carbon dioxide and obtains 168.4 kg of urea. Determine the limiting reactant, theoretical yield of urea, and percent yield for the reaction.

Video transcript

Hey everyone, we're told that the reaction of ammonia with carbon dioxide produces Yuria a common fertilizer and were provided the reaction below note the reaction is unbalanced. A chemist uses 32.5 mg of ammonia with 89.6 mg of carbon dioxide to produce 55.7 mg of Yuria. In industrial Yuria synthesis, which is the limiting reactant. And what is the theoretical yield of Yuria, calculate the percent yield for the reaction first. Let's go ahead and balance out this reaction. Now, in order to balance this out, all we need to do is add a coefficient of two prior to our ammonia and everything will be completely balanced out. Now let's go ahead and determine our limiting reagent. Starting with our 32.5 mg of ammonia, we want to convert this into a mole of Syria. So, using dimensional analysis, we're going to convert our milligrams into grams and we know that we have 10 to the third milligrams per one g. Now, using the molar mass of Ammonia, we know that we have 17.031g of ammonia Per one Mole of Ammonia. Now looking at our multiple ratios between ammonia and urea, We see that we have two mol of ammonia per one mole of iria. And this can be seen in our reaction that we just balanced out. So when we calculate this out and cancel out all of our units, We end up with 9.541, 4 times 10 to the -4 mol of Iria. Now let's go ahead and take our 89. mg of carbon dioxide and convert this into mole of your area as well. Again, using dimensional analysis, we know that we have 10 to the third milligrams per one g. And using carbon dioxides Mueller mass, we know that we have 44.01 g of carbon dioxide per one mole of carbon dioxide. Looking at our multiple ratios, we see that per one mole of carbon dioxide, we have one mole of iria. Now, when we calculate this out and cancel out all of our units, we end up with 2.359 times to the negative three mole of syria. Now, comparing these two values, we see that the least amount of moles produced of Yuria comes from our ammonia. So this means that ammonia is our limiting re agent. Now that we've determined our limiting reagent, we're going to take that value and determine our theoretical yield. So we have 9.5414 times 10 to the negative four mol of iria. And we're going to convert this into milligrams of Yuria. Looking at areas Mueller mass, we know that we have 64.80 g of Yuria per one mole of Yuria, converting our grams into milligrams, we know that we have 10 to the third milligrams per one g. Now, when we calculate this out, we end up with a value of 61. mg of Iria and this is going to be our theoretical yield. Now let's go ahead and determine our percent yield. To determine our percent yield. We're going to take our actual, which was said to be 55.75 mg of iria and divided by our theoretical, which we calculated to be 61. mg of iria. And since we want this in percentage, we're going to multiply this by 100%. So this gets us to a value of 91.10% and this is going to be our percent yield. Now, I hope this made sense and let us know if you have any questions.